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MATHS 5 Marks Questions

Co-ordinate Geometry · UP Board (UPMSP) STD 10 (UPMSP - English Medium). 81 questions.

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Question 15 Marks
The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex.
Answer
Let A(3, 4), B(3, 8) and C(9, 8) be the three given vertex then fourth vertex be D(x, y)
Since, ABCD is parallelogram, the diagonals bisect each other.
Therefore, the mid-point of the diagonals of the parallelogram coincide.
Let P(x, y) be the mid-point of diagonal AC then,
$\text{P}(\text{x},\text{y})=\Big(\frac{3+9}{2},\frac{4+8}{2}\Big)$ P(x, y) = (6, 6)Let Q(x, y) be the mid-point of diagonal BD then,
$\text{Q}(\text{x},\text{y})=\Big(\frac{3+\text{x}}{2},\frac{8+\text{y}}{2}\Big)$ Coordinates of mid-point AC = Coordinates of mid-point BD P(x, y) = Q(x, y) $\Rightarrow\ (6,6)=\Big(\frac{3+\text{x}}{2},\frac{8+\text{y}}{2}\Big)$ Now, equating individual components, $\Rightarrow\ 6=\frac{3+\text{x}}{2}$ and $6=\frac{8+\text{y}}{2}$ ⇒ 3 + x = 12 and 8 + y = 12 ⇒ x = 9 and y = 4Hence, coordinates of fourth point are (9, 4).
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Question 25 Marks
The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of $\triangle\text{ABC}.$
Answer
Given that, the points $A(2,9), B(a, 5)$ and $C(5,5)$ are the vertices of a $\triangle A B C$ right angled at $B$. By pythagoras theorem,
$AC ^2= AB ^2+ BC ^2 \ldots .$. .(i)
Now, by distance formula, $AB =\sqrt{( a -2)^2+(5-9)^2}$
$[\because$ distance between two points
$\left( x _1, y _1\right)$ and $\left.\left( x _2, y _2\right)=\sqrt{\left( x _2- x _1\right)^2+\left( y _2- y _1\right)^2}\right]=\sqrt{ a ^2+4-4 a +16}=\sqrt{ a ^2-4 a +20}$
$BC =\sqrt{(5- a )^2+(5-5)^2}=\sqrt{(5- a )^2+0}=5- a$ and $AC =\sqrt{(2-5)^2+(9-5)^2}$
$=\sqrt{(-3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5$
Put the values of $A B, B C$ and $A C$ in Eq. (i),
we get $(5)^2=\sqrt{\left(a^2-4 a+20\right)^2}+(5-a)^2 $
$\Rightarrow 25=a^2-4 a+20+25+a^2-10 a$
$ \Rightarrow 2 a^2-14 a+20=0 $
$\Rightarrow a^2-7 a+10= 0$
$\Rightarrow a^2-2 a-5 a+10=0[$ by factorisation method$]$
 $\Rightarrow a(a-2)-5(a-2)=0 $
$\Rightarrow(a-2)(a-5)=0 $
$\therefore a=2,5$ Here,
$a \neq 5$, since at $a=5$,
the length of $B C=0$.
It is not possible because the sides $A B, B C$ and $C A$ from a right angled triangle.
So, $a=2$ Now, the coordinate of $A, B$ and $C$ becomes $(2,9),(2,5)$ and $(5,5)$ respectively.
$\therefore \text { Area of } \triangle ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$
$\therefore \triangle=\frac{1}{2}[2(5-5)+2(5-9)+5(9-5)]$
$=\frac{1}{2}[2 \times 0+2(-4)+5(4)]$
$=\frac{1}{2}(0-8+20)=\frac{1}{2} \times 12=6$
Hence, the required area of $\triangle ABC$ is 6 sq . units.
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Question 35 Marks
Find the lengths of the medians of a triangle whose vertices are A(-1, 3), B(1, -1) and C(5, 1).
Answer

Let AD, BF and CE be the medians of $\triangle\text{ABCD}$
Coordinates of D are $\Big(\frac{5+1}{2},\frac{1-1}{2}\Big)=(3,0)$
Coordinates of E are $\Big(\frac{-1+1}{2},\frac{3-1}{2}\Big)=(0,1)$
Coordinates of F are $\Big(\frac{5-1}{2},\frac{1+3}{2}\Big)=(2,2)$
Length of $\text{AD}=\sqrt{(-1-3)^2+(3-0)^2}=5\text{ units}$
Length of $\text{BF}=\sqrt{(2-1)^2+(2+1)^2}=\sqrt{10}\text{ units}$
Length of $\text{CE}=\sqrt{(5-0)^2+(1-1)^2}=5\text{ units}$
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Question 45 Marks
Find the ratio in which the point (2, y) divides the line segment joining the points A(-2, 2) and B(3, 7). Also, find the value of y.
Answer
Let the point P(2, y) divide the line segment joining the points A(-2, 2) and B(3, 7) in the ratio k : 1
Then, the coordinates of P are,
$\bigg[\frac{3\text{k}+(-2)\times1}{\text{k+1}},\frac{7\text{k}+2\times1}{\text{k+1}}\bigg]$
$\bigg[\frac{3\text{k}-2}{\text{k+1}},\frac{7\text{k}+2}{\text{k}+1}\bigg]$
But the coordinates of P are given as (2, y).
$\therefore\ \frac{3\text{k}-2}{\text{k}+1}=2$
$\Rightarrow\ 3\text{k}-2=2\text{k}+2$
$\Rightarrow\ 3\text{k}-2\text{k}=2+2$
$\Rightarrow\ \text{k}=4$
$\frac{7\text{k}+2}{\text{k}+1}=\text{y}$
Putting the value of k, we get
$\frac{7\times4+2}{4+1}=\text{y}$
$\frac{30}{5}=\text{y}$
$6=\text{y}$
i.e., $\text{y}=6$
Hence, the ratio is 4 : 1 and y = 6.
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Question 55 Marks
Prove that the points (-4, -1), (-2, 4), (4, 0) and (2, 3) are the vertices of a rectangle.
Answer
Let the vertices of a quadrilateral ABCD are A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) Join AC and BD which intersect each other at O. If O is the mid-point of AC then its coordinates will be $=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)$ or $\Big(\frac{-4+4}{2},\frac{-1+0}{2}\Big)$ $=\Big(\frac{0}{2},\frac{-1}{2}\Big)$ or $\Big(0,\frac{-1}{2}\Big)$If O is the mid-point of BD, then its co-ordinates will be $=\Big(\frac{-2+2}{2},\frac{-4+3}{2}\Big)$
or $\Big(\frac{0}{2},\frac{-1}{2}\Big)$ or $\Big(0,\frac{-1}{2}\Big)$ $\because$ The mid-points of AC and BD are the same. $\therefore$ AC and AD bisect eachother at O Now, $\text{AC}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$ $=\sqrt{[4-(-4)]^2+[0-(-1)]^2}$ $=\sqrt{(4+4)^2+(1)^2}=\sqrt{(8)^2+(1)^2}$ $=\sqrt{64+1}=\sqrt{65}$ and $\text{BD}=\sqrt{[2-(-2)]^2+[3-(-4)]^2}$ $=\sqrt{(2+2)^2+(3+4)^2}=\sqrt{(4)^2+(7)^2}$ $=\sqrt{16+49}=\sqrt{65}$ $\because$ Diagonal bisect eachother at O and are equal. $\therefore$ ABCD is a rectangle.
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Question 65 Marks
If the points A(1, -2), B(2, 3), C(a, 2) and D(-4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.
Answer
In parallelogram, we know that, diagonals bisects each other i.e., mid-point of AC = mid-point of BD
$\Rightarrow\left(\frac{1+ a }{2}, \frac{-2+2}{2}\right)=\left(\frac{2-4}{2}, \frac{3-3}{2}\right) \Rightarrow \frac{1+ a }{2}=\frac{2-4}{2}=\frac{-2}{2}=-1 \quad$ Since, mid-point of a line segment having points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\left.\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\right] \Rightarrow 1+a=-2 \Rightarrow a=-3$ So, the required value of a is -3 . Given that, $A B$ as base of a parallelogram and drawn a perpendicular from $D$ to $A B$ which meet $A B$ at $P$. So, $D P$ is a height of a parallelogram. Now, equation of base $A B$, passing through the points $(1,-2)$ and $(2,3)$ is $\Rightarrow\left( y - y _1\right)=\frac{ y _2- y _1}{ x _2- x _1}\left( x - x _1\right) \Rightarrow( y +2)=\frac{3+2}{2-1}( x -1) \Rightarrow( y +2)=5( x -1) \Rightarrow 5 x - y =7$ (i) Slope of $A B$, say $m _1=\frac{ y _2- y _1}{ x _2- x _1}=\frac{3+2}{2-1}=5$ Let the slope of DP be $m _2$. Since, DP is perpendicular to $A B$. By condition of perpendicularity $m _1 \times m _2=-1 \Rightarrow 5 m_2=-1 \Rightarrow m_2=-\frac{1}{5}$ Now, Eq. of DP, having slope ( $-\frac{1}{5}$ ) and passing the point $(-4,-3)$ is $\left(y-y_1\right)=m_2\left(x-x_1\right) \Rightarrow(y+3)=-\frac{1}{5}(x+4) \Rightarrow 5 y+15=-x-4 \Rightarrow x+5 y=-19$ (ii) On adding Eq. (i) and (ii), then we get the intersection point P. Put the value of $y$ from Eq. (i) in Eq. (ii), we get $x+5(5 x-7)=-19$ [using Eq. (i)] $\Rightarrow x+25 x-35=-19 \Rightarrow 26 x=16 \therefore x=\frac{8}{13}$ Put the value of $x$ in Eq. (i), we get $y=5\left(\frac{8}{13}\right)-7=\frac{40}{13}-7 \Rightarrow y=\frac{40-91}{13} \Rightarrow y=\frac{-51}{13}$
$\therefore$ Coordinates of point $P =\left(\frac{8}{13}, \frac{-51}{13}\right)$
So, length of the height of a parallelogram, $DP =\sqrt{\left(\frac{8}{13}+4\right)^2+\left(\frac{-51}{13}+3\right)^2}[\because$ by distance formula, distance two points $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$ is $\left.d =\sqrt{\left( x _2- x _1\right)^2\left( y _2- y _1\right)^2}\right] \Rightarrow DP =\sqrt{\left(\frac{60}{13}\right)^2+\left(\frac{-12}{13}\right)^2}$ $=\frac{1}{13} \sqrt{3600+144}=\frac{1}{13} \sqrt{3744}=\frac{12 \sqrt{26}}{13}$ Hence, the required length of height of a parallelogram is $\frac{12 \sqrt{26}}{13}$.
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Question 75 Marks
The points (3, -4) and (-6, 2) are the extremities of a diagonal of a parallelogram. If the third vertex is (-1, -3). Find the coordinates of the fourth vertex.
Answer

Let A(3, -4) and C(-6, -2) be the extremities of diagonal AC and B(-1, -3), D(x, y) be the extremities of diagonal BD.
Since the diagonals of a parallelogram bisect each other.
$\therefore$ Coordinates of mid-point of AC = Coordinates of mid-point of BD.
$\Rightarrow\ \frac{3-6}{2}=\frac{\text{x}-1}{2}$
$\Rightarrow\ \frac{-3}{2}=\frac{\text{x}-1}{2}$
$\Rightarrow\ \text{x}=-2$
And, $\frac{-4+2}{2}=\frac{\text{y}-3}{2}$
$\Rightarrow\ \frac{-2}{2}=\frac{\text{y}-3}{2}$
$\Rightarrow\ \text{y}=1$
Hence, fourth vertex of parallelogram is (-2, 1).
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Question 85 Marks
The centre of a circle is (2a, a - 7). Find the values of a if the circle passes through the point (11, -9) and has diameter $10\sqrt{2}$ units.
Answer
By given condition, Distance between the centre C (2a, a - 7) and the point P (11, -9), which lie on the circle = Radius of circle
$\therefore \text { Radius of circle }=\sqrt{(11-2 a)^2+(-9-a+7)^2} \ldots \ldots \text { (i) }\left[\because \text { distance between two points }\left(x_1, y_1\right)\right. \text { and }$
$\left.\left(x_2, y_2\right)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]$
Given that, length of diameter $=10 \sqrt{2} \therefore$ Length of radius $=\frac{\text { Lenght of diameter }}{2}$
$=\frac{10 \sqrt{2}}{2}=5 \sqrt{2}$
Put this value in eq.(i), we get $5 \sqrt{2}=\sqrt{(11-2 a )^2+(-2- a )^2}$
Squaring on both sides, we get
$50=(11-2 a)^2+(2+a)^2$
$\Rightarrow 50=121+4 a^2-44 a+4+a^2+4 a$
$\Rightarrow 5 a^2-40 a+75=0$
$\Rightarrow a^2-8 a+15=0$
$\Rightarrow a^2-5 a-3 a+15=0$
[By factorisation method] $\Rightarrow a(a-5)-3(a-5)=0$
$\Rightarrow(a-5)(a-3)=0$
$\therefore a=3,5$ Hence, the required values of $a$ are 5 and 3 .
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Question 95 Marks
Find the coordinates of the points which divide the line segment joining the points (-4, 0) and (0, 6) in four equal parts.
Answer
The co-ordinates of the midpoint $\left(x_m, y_m\right)$ between two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is given by,
$\left(x_{m}, y_{m}\right)=\left[\left(\frac{x_1+x_2}{2}\right),\left(\frac{y_1+y_2}{2}\right)\right]$
Here we are supposed to find the points which divide the line joining $A(-4,0)$ and $B(0,6)$ into 4 equal parts. We shall first find the midpoint $M(x, y)$ of these two points since this point will divide the line into two equal parts, $\left( x _{ m }, y _{ m }\right)=\left[\left(\frac{-4+0}{2}\right),\left(\frac{0+6}{2}\right)\right]\left( x _{ m }, y _{ m }\right)=(-2,3)$ So the point $M (-2,3)$ splits this line into two equal parts. Now, we need to find the midpoint of $A(-4,0)$ and $M(-2,3)$ separately and the midpoint of $B(0,6)$ and $M(-2,3)$. These two points along with $M(-2,3)$ split the line joining the original two points into four equal parts. Let $M_1(e, d)$ be the midpoint of $A(-4,0)$ and $M(-2,3) .(e, d)=\left[\left(\frac{-4-2}{2}\right),\left(\frac{0+3}{2}\right)\right](e, d)=\left(-3, \frac{3}{2}\right)$ Now let $M_2(g, h )$ be the midpoint of $B(0,6)$ and $M(-2,3) \cdot(g, h)=\left[\left(\frac{0-2}{2}\right),\left(\frac{6+3}{2}\right)\right]( g , h )=\left(-1, \frac{9}{2}\right)$ Hence the co-ordinates of the points which divide the line joining the two given points are $\left(-3, \frac{3}{2}\right),(-2,3)$ and $\left(-1, \frac{9}{2}\right)$.
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Question 105 Marks
Find the ratio in which the line segment joining (-2, -3) and (5, 6) is divided by (i) x-axis (ii) y-axis. Also, find the coordinates of the point of division in each case.
Answer
  1. Let x-axis divides PQ in the ratio.
$\lambda:1$

Let R(x, y) be the coordinates of the point of division.

Then, the coordinates of the point of division are,



$\text{R}\bigg(\frac{5\times\lambda+(-2)\times1}{\lambda+1},\frac{6\times\lambda+(-3)\times1}{\lambda+1}\bigg)=\bigg(\frac{5\lambda-2}{\lambda+1},\frac{6\lambda-3}{\lambda+1}\bigg)$

Since R lies on x-axis.

Therefore, y-coordinates of every point on x-axis is zero.

$\therefore\ \frac{6\lambda-3}{\lambda+1}=0$

$\Rightarrow\ 6\lambda-3=0$

$\Rightarrow\ 6\lambda=3$

$\Rightarrow\ \lambda=\frac{3}{6}=\frac{1}{2}$

Hence, the required ratio is $\frac{1}{2}:1$ or 1 : 2

Let point R(x, y) divide the line joining in the ratio 1 : 2.

Putting $\lambda=\frac{1}{2}$ in the coordinates of R, we get

$\text{R}(\text{x, y})=\Bigg[\Bigg(\frac{5\times\frac{1}{2}-2}{\frac{1}{2}+1}\Bigg)\Bigg(\frac{6\times\frac{1}{2}-3}{\frac{1}{2}+1}\Bigg)\Bigg]$

Now, equating the individual components,

$\Rightarrow\ \text{x}=\frac{\frac{5-2}{2}}{\frac{1+1}{2}}$ and $\text{y}=\frac{3-3}{\frac{1+2}{1}}$

$\Rightarrow\ \text{x}=\frac{1}{3}$ and $\text{y}=0$

$\text{R}(\text{x,y})=\Big(\frac{1}{3},0\Big)$

Hence, the point $\Big(\frac{1}{3},0\Big)$ divide the line joining PQ in the ratio 1 : 2.
  1. $\because$ Abscissa of a point on y-axis is zero.
Let the point be (0, y).

Let this point divides the line segment joining (-2, -3) and (5, 6) in the ratio m : n.

$\therefore\ 0=\frac{\text{mx}_2+\text{nx}_1}{\text{m+n}}=\frac{\text{m(5)+n(-2)}}{\text{m+n}}$

$\Rightarrow\ 0=\frac{5\text{m}-2\text{n}}{\text{m+n}}\Rightarrow5\text{m}-2\text{n}=0$

$\Rightarrow\ 5\text{m}=2\text{n}\Rightarrow\ \frac{\text{m}}{\text{n}}=\frac{2}{5}$

$\therefore$ Ratio = 2 : 5

and $\text{y}=\frac{2\times6+5\times(-3)}{2+5}=\frac{12-15}{7}=\frac{-3}{7}$

$\therefore$ Required point $=\Big(0,\frac{-3}{7}\Big)$
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Question 115 Marks
Prove that (4, 3), (6, 4) (5, 6) and (3, 5) are the angular points of a square.
Answer
Let the given points be, A(4, 3), B(6, 4) C(5, 6) and D(3, 5) respectively. Then, Now, $\text{AB}=\sqrt{(6-4)^2+(4-3)^2}$ $=\sqrt{4+1}$ $=\sqrt{5}\ \text{units}$ and $\text{BC}=\sqrt{(5-6)^2+(6-4)^2}$$=\sqrt{1+4}$
$=\sqrt{5}\ \text{units}$ $\text{CD}=\sqrt{(3-5)^2+(5-6)^2}$ $=\sqrt{4+1}$ $=\sqrt{5}\ \text{units}$ $\text{AD}=\sqrt{(3-4)^2+(5-3)^2}$ $=\sqrt{1+4}$ $=\sqrt{5}\ \text{units}$ Thus, AB = BC = CD = AD Diagonal $\text{AC}=\sqrt{(5-4)^2+(6-3)^2}$ $=\sqrt{1+9}$ $=\sqrt{10}\ \text{units}$ Diagonal $\text{BD}=\sqrt{(3-6)^2+(5-4)^2}$ $=\sqrt{9+1}$ $=\sqrt{10}\ \text{units}$ $\therefore$ AB = BC = CD = AD and diagonal AC = diagonal BD. Thus, ABCD is a quadrilateral in which all sides are equal and the diagonal are equal. Hence, ABCD is a square.
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Question 125 Marks
If P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.
Answer
P(-5, -3), Q(-4, -6), R(2, -3) and S(1,2) are the vertices of a quadrilateral PQRS. Join PR which forms two triangles PQR and PSR,
Now area of $\triangle\text{PQR}$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[-5(-6+3)+(-4)(-3+3)+2(-3+6)$
$=\frac{1}{2}[-5\times(-3)+(-4\times0)+2\times3]$
$=\frac{1}{2}(15+0+6)=\frac{1}{2}\times21=\frac{21}{2}$
and area of $\triangle\text{PSR}$
$=\frac{1}{2}[-5\times(2+3)+1(-3+3)+2(-3-2)]$
$=\frac{1}{2}[-5\times5+1\times0+2\times(-5)]$
$=\frac{1}{2}[-25+0-10]=\frac{1}{2}\times(-35)=\frac{-35}{2}$
$\therefore$ Area of quadrilateral PQRS $=\frac{21}{2}+\frac{35}{2}=\frac{56}{2}=28\text{ sq units}$
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Question 135 Marks
Find the area of a triangle whose vertices are,
$(\text{at}_1^2, 2\text{at}_1), (\text{at}_2^2, 2\text{at}_2)$ and $(\text{at}_3^2, 2\text{at}_3).$
Answer
Co-ordinates of $\triangle\text{ABC}$ are $\text{A}(\text{at}_1^2, 2\text{at}_1), \text{B}(\text{at}_2^2, 2\text{at}_2)$ and $\text{C}(\text{at}_3^2, 2\text{at}_3)$$\therefore$ Area of $\triangle=\frac{1}{2}|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)|$
$=\frac{1}{2}|\text{at}_1^2(2\text{at}_2-2\text{at}_3)+\text{at}_2^2(2\text{at}_3-2\text{at}_1)+\text{at}_3^2(2\text{at}_1-2\text{at}_2)|$
$=\frac{1}{2}|2\text{a}^2\text{t}_1^2\text{t}_2-2\text{a}^2\text{t}_1^2\text{t}_3+2\text{a}^2\text{t}_2^2\text{t}_3-2\text{a}^2\text{t}_2^2\text{t}_1+2\text{a}^2\text{t}_3^2\text{t}_1-2\text{a}^2\text{t}_3^2\text{t}_2|$
$=\frac{1}{2}\times2\text{a}^2|\text{t}_1^2\text{t}_2-\text{t}_1^2\text{t}_3+\text{t}_2^2\text{t}_3-\text{t}_2^2\text{t}_1+\text{t}_3^2\text{t}_1-\text{t}_3^2\text{t}_2|$
$= \text{a}^2|\text{t}_1^2\text{t}_2 - \text{t}_1^2\text{t}_3 + \text{t}_2^2\text{t}_3 - \text{t}_3^2\text{t}_2 -\text{t}_2^2\text{t}_1 + \text{t}_3^2\text{t}_1|$
$= \text{a}^2|\text{t}_1^2(\text{t}_2 - \text{t}_3) + \text{t}_2\cdot\text{t}_3(\text{t}_2 - \text{t}_3) - \text{t}_1(\text{t}_2^2 -\text{ t}_3^2)|$
$= \text{a}^2|\text{t}_1^2(\text{t}_2 - \text{t}_3) + \text{t}_2\text{t}_3(\text{t}_2 - \text{t}_3) - \text{t}_1(\text{t}_2 +\text{ t}_3)(\text{t}_2-\text{t}_3)|$
$= \text{a}^2|(\text{t}_2 - \text{t}_3)[\text{t}_1^2 +\text{t}_2\text{t}_3 - \text{t}_1\text{t}_2 - \text{t}_1\text{t}_3]|$
$= \text{a}^2|(\text{t}_2 - \text{t}_3)[\text{t}_1^2 -\text{t}_1\text{t}_2 - \text{t}_1\text{t}_3 +\text{t}_2\text{t}_3]|$
$= \text{a}^2|(\text{t}_2 - \text{t}_3)[\text{t}_1(\text{t}_1 - \text{t}_2) - \text{t}_3(\text{t}_1 - \text{t}_2)]|$
$= \text{a}^2|(\text{t}_2 - \text{t}_3)(\text{t}_1 - \text{t}_2)(\text{t}_1 - \text{t}_3)|$
$= \text{a}^2(\text{t}_1 - \text{t}_2)(\text{t}_2 - \text{t}_3)(\text{t}_3 - \text{t}_1)$
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Question 145 Marks
Prove that the points (3, 0), (6, 4) and (-1, 3) are vertices of a right-angled isosceles triangle.
Answer


Let A(3, 0), B(6, 4) and C(-1, 3) be the given points. Then,
$\text{AB}=\sqrt{(6-3)^2+(4-0)^2}$
$=\sqrt{9+16}=5\text{ units}$
$\text{BC}=\sqrt{(-1-6)^2+(3-4)^2}$
$=\sqrt{49+1}=\sqrt{50}\text{ units}$
$\text{AC}=\sqrt{(-1-3)^2+(3-0)^2}$
$=\sqrt{16+9}=5\text{ units}$
Thus, $A B=A C=5$ units
$\therefore \triangle ABC$ is isosceles.
Also, $A B^2+A C^2=5^2+5^2=50$ and
$BC ^2=(\sqrt{50})^2=50$
Thus, $A B^2+ AC ^2= BC ^2$
This shows that $\triangle ABC$ is right-angled at A .
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Question 155 Marks
The points A(2, 0), B(9, 1) C(11, 6) and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.
Answer
The co-ordinates of vertices of a quadrilateral ABCD are A(2, 0), B(9, 1), C(11, 6) and D(4, 4) AC and BD are its diagonals which intersect each other at O. Let O is the mid-point of AC then co-ordinates of O will be $\Big(\frac{2+11}{2},\frac{0+6}{2}\Big)$ or $\Big(\frac{13}{2},\frac{6}{2}\Big)$ or $\Big(\frac{13}{2},3\Big)$ Let O is the mid-point of BD, then co-ordinates of O will be $\Big(\frac{9+4}{2},\frac{1+4}{2}\Big)$ or $\Big(\frac{13}{2},\frac{5}{2}\Big)$The co-ordinates of O in both cases are not same.
It is not a parallelogram and also not a rhombus.
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Question 165 Marks
Three consecutive vertices of a parallelogram are (-2, 1), (1, 0) and (4, 3). Find the fourth vertex.
Answer
Let the co-ordinates of three vertices are A(-2, -1), B(1, 0) and C(4, 3) and let the diagonals AC and BD bisect each other at O. $\because$ O is the mid-point of AC. $\therefore$ Vertices of O will be$\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)$ or $\Big(\frac{-2+4}{2},\frac{-1+3}{2}\Big)$
or $\Big(\frac{2}{2},\frac{2}{2}\Big)$ or (1, 1)
Let co-ordinates of the fourth vertex D be (x, y)
$\because$ O is the mid-point of BD $\therefore$ co-ordinates of O will be $\Big(\frac{1+\text{x}}{2},\frac{0+\text{y}}{2}\Big)$ or $\Big(\frac{1+\text{x}}{2},\frac{\text{y}}{2}\Big)$ $\therefore\ \frac{1+\text{x}}{2}=1$ $\Rightarrow\ 1+\text{x}=2\Rightarrow\ \text{x}=2-1=1$ and $\frac{\text{y}}{2}=1\Rightarrow\ \text{y}=2$ Co-ordinates of D will be (1, 2).
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Question 175 Marks
The vertices of ΔABC are (-2, 1), (5, 4) and (2, -3) respectively. Find the area of the triangle and the length of the altitude through A.
Answer
Given: The vertices of triangle $A B C$ are $A(-2,1)$ and $B(5,4)$ and $C(2,-3)$ To find: The area of triangle $A B C$ and length if the altitude through A.
Proof: We know area of triangle formed by three points $\left( x _1, y _1\right),\left( x _2, y _2\right)$ and $\left( x _3, y _3\right)$ is given by
$\Delta=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
Now Area of $\triangle ABC$
Taking three points $A (-2,1)$ and $B (5,4)$ and $C (2,-3)$
Area $(\triangle ABC )=\frac{1}{2}|(-8-15+2)-(5+8+6)|$
$=\frac{1}{2}|(-21)-(19)|$
$=\frac{1}{2}|(-40)|$
$=\frac{1}{2}(40)$
$=20$
We have, $BC =\sqrt{(5-2)^2+(4+3)^2}$
$BC=\sqrt{(3)^2+(7)^2}$
$BC=\sqrt{9+49}$
$BC=\sqrt{58}$
Now, Area $(\triangle ABC )=\frac{1}{2} \times BC \times$ length of altitude through A
$20=\frac{1}{2} \times \sqrt{58} \times$ length of altitude through $A$
length of altitude through $A =\frac{40}{\sqrt{58}}$
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Question 185 Marks
The point A divides the join of P(-5, 1) and Q(3, 5) in the ratio k : 1. Find the two values of k for which the area of ΔABC where B is (1, 5) and C(7, -2) is equal to 2 units.
Answer
Given that,
Point A divides the line segment joining P(-5, 1) and Q(3, 5) in the ratio k : 1.
So the coordinates of A are,
$\Big(\frac{3\text{k}-5}{\text{k}+1},\frac{5\text{k}+1}{\text{k}+1}\Big)$
We know that, area of triangle formed by three points $\left( x _1, y _1\right),\left( x _2, y _2\right)$ and $\left( x _3, y _3\right)$ is given by.
$\triangle=\frac{1}{2}|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)|$
It is given that area of $\triangle\text{ABC}$ is 2 square units.
Now, three points are
$\text{A}\Big(\frac{3\text{k}-5}{\text{k}+1},\frac{5\text{k}+1}{\text{k}+1}\Big),$ B(1, 5) and C(7, -2)
$\Rightarrow\ 2=\frac{1}{2}\Big|\frac{3\text{k}-5}{\text{k}+1}(5-(-2))+1\Big[(-2)-\Big(\frac{5\text{k}+1}{\text{k}+1}\Big)\Big]+7\Big(\frac{5\text{k}+1}{\text{k}+1}-5\Big)\Big|$
$\Rightarrow\ 2=\frac{1}{2}\Big|\frac{21\text{k}-35}{\text{k}+1}+\Big(\frac{-2\text{k}-2-5\text{k}-1}{\text{k}+1}\Big)+\Big(\frac{35\text{k}+7-35\text{k}-35}{\text{k}+1}\Big)\Big|$
$\Rightarrow\ 2=\frac{1}{2}\Big|\frac{21\text{k}-35-7\text{k}-3-28}{\text{k}+1}\Big|$
$\Rightarrow\ 2=\frac{1}{2}\Big|\frac{14\text{k}-66}{\text{k}+1}\Big|$
$\Rightarrow\ 2=\frac{1}{2}\times2\Big|\frac{7\text{k}-33}{\text{k}+1}\Big|$
$\Rightarrow\ \frac{7\text{k}-33}{\text{k}+1}=\pm2$
$\Rightarrow\ 7\text{k}-33=\pm2(\text{k}+1)$
⇒ 7k - 33 = 2k + 2 or 7k - 33 = -2k - 2
⇒ 5k = 35 or 9k = -2 + 33
⇒ k = 7 or $\text{k}=\frac{31}{9}$
Hence, the value of k is either 7 or $\frac{31}{9}.$
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Question 195 Marks
Find the area of a quadrilateral ABCD, the coordinates of whose varities are A(-3, 2), B(5, 4), C(7, -6) and D(-5, -4).
Answer
Area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD


$=\frac{1}{2}|(-12-30+14)-(10+28+18)|$$+\frac{1}{2}(18-28-10)-(14+30+12)|$
$=\frac{1}{2}|-28-56|+\frac{1}{2}|-20-56|$
$=\frac{1}{2}|-84|+\frac{1}{2}|-76|$
$=42+38=80\text{ sq.units}$
$\therefore$ Area of quadrilateral = 80 sq.units
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Question 205 Marks
The base PQ of two equilateral triangles PQR and PQR' with side 2a lies along y-axis such that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R' of the triangles.
Answer
∆PQR and PQR’ are equilateral triangles with side 2a each and base PQ and mid of point of PQ is O(0, 0) and PQ lies along y-axis,
 
$\because P R=Q R=P R^{\prime}=Q R^{\prime}=2 a P O=O Q=a \therefore \text { In right } \triangle P R O$
$P R^2=P O^2+O R^2(P y t h a g o r a s \text { Theorem }) \Rightarrow(2 a)^2=(a)^2+O R^2 \Rightarrow 4 a^2=a^2+O R^2 O R^2=4 a^2-a^2=3 a^2$
$\therefore O R=\sqrt{3} a$
Similarly $OR ^{\prime}=-\sqrt{3} a$, Now, co-ordinates of $R$ will be $(\sqrt{3} a , 0)$ and co-ordinates of $R$ will be $(-\sqrt{3} a , 0)$
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Question 215 Marks
The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the third vertex.
Answer
Given: The area of triangle is 5 . Two of its vertices are $(2,1)$ and $(3,-2)$. The third vertex lies on $y=x+3$.
To find: The third vertex.
Proof: Let the third vertex be ( $x , y$ )
We know area of triangle formed by three points $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is given by,
$\triangle=\frac{1}{2}|(\text{x}_1\text{y}_2+\text{x}_2\text{y}_3+\text{x}_3\text{y}_1)-(\text{x}_2\text{y}_1+\text{x}_3\text{y}_2+\text{x}_1\text{y}_3)|$
Now,
Taking three points (x, y), (2, 1) and (3, -2)
$\triangle=\frac{1}{2}|(\text{x}-4+3\text{y})-(2\text{y}+3-2\text{x})|$
$\triangle=\frac{1}{2}|3\text{x}+\text{y}-7|$
$5=\frac{1}{2}|3\text{x}+\text{y}-7|$
$\pm10=3\text{x}+\text{y}-7$
$10 = 3x + y - 7 or -10 = 3x + y - 7$
$0 = 3x + y - 17........(i) or 0 = 3x + y + 3 ......(ii)$
Also, it is given the third vertex lies on y = x + 3
Substituting the value in equation (i) and (ii) we get
$\pm10=3\text{x}+\text{y}-7$
$10 = 3x + y - 7$
$0 = 3x + y - 17........(i)$
$0 = 3x + (x + 3) - 17$
$\text{x}=\frac{7}{2}$
Again substituting the value of x in equation (i) we get
0 = 3x + y - 17........(i)
$0=3\Big(\frac{7}{2}\Big)+\text{y}-17$
$\text{y}=\frac{13}{2}$
Hence, $\Big(\frac{7}{2},\frac{13}{2}\Big)$
Similarly,
$-10 = 3x + y - 7$
$0 = 3x + y + 3 ......(ii)$
$0 = 3x + (x + 3) + 3$
$\text{x}=\frac{-3}{2}$
Again substituting the value of x in equation (ii) we get
$0 = 3x + y + 3 ......(ii)$
$0=3\Big(\frac{-3}{2}\Big)+\text{y}+3$
$\text{y}=\frac{3}{2}$
Hence, $\Big(\frac{-3}{2},\frac{3}{2}\Big)$
Hence the coordinates of $\Big(\frac{7}{2},\frac{13}{2}\Big)$ and $\Big(\frac{-3}{2},\frac{3}{2}\Big).$
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Question 225 Marks
The coordinates of the point P are (-3, 2). Find the coordinates of the point Q which lies on the line joining P and origin such that OP = OQ.
Answer
Co-ordinates of P are (-3, 2) and origin O are (0, 0)
Let co-ordinates of Q be (x, y)
O is the mid point of PQ

O is the mid point so that
$\text{O}=\frac{-3+\text{x}}{2}$
$\text{O}=-3+\text{x}$
$\text{x}=3$
$\text{O}=\frac{2+\text{y}}{2}$
$\text{O}=2+\text{y}$
$\text{y}=-2$
Therefore Co-ordinate of Q(x, y) = (3, -2).
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Question 235 Marks
Prove that the points (4, 5) (7, 6), (6, 3), (3, 2) are the vertices of a parallelogram. Is it a rectangle.
Answer
Let $A(4,5) ; B(7,6) ; C(6,3)$ and $D(3,2)$ be the vertices of a quadrilateral. We have to prove that the quadrilateral $A B C D$ is a parallelogram. We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram. Now to find the mid-point $P(x, y)$ of two points $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ we use section
,$\text{P(x, y)}=\bigg(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\bigg)$
So the mid-point of the diagonal AC is,
$\text{Q(x, y)}=\bigg(\frac{4+6}{2},\frac{5+3}{2}\bigg)$ = (5, 4) Similarly mid-point of diagonal BD is, $\text{R(x,y)}=\bigg(\frac{7+3}{2},\frac{6+2}{2}\bigg)$ = (5, 4) Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other. Hence ABCD is a parallelogram. Now to check if ABCD is a rectangle, we should check the diagonal length. $\text{AC}=\sqrt{(6-4)^2+(3-5)^2}$ $=\sqrt{4+4}$ $=2\sqrt{2}$ Similarly, $\text{BD}=\sqrt{(7-3)^2+(6-2)^2}$ $=\sqrt{16+16}$ $=4\sqrt{2}$ Diagonals are of different lengths. Hence ABCD is not a rectangle.
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Question 245 Marks
The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.
Answer
The distance d between two points $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$ is given by the formula, $d =\sqrt{\left( x _1- x _2\right)^2+\left( y _1- y _2\right)^2}$ Here it is given that one end of a line segment has co-ordinates $(2,-3)$. The abscissa of the other end of the line segment is given to be 10. Let the ordinate of this point be ' $y$ '.
So, the co-ordinates of the other end of the line segment is $(10, y)$.
The distance between these two points is given to be 10 units.
Substituting these values in the formula for distance between two points we have,
$d =\sqrt{(2-10)^2+(-3- y )^2} 10=\sqrt{(-8)^2+(-3- y )^2}$ Squaring on both sides of the equation we have,
$100=(-8)^2+(-3-y)^2$
$100=64+9+y^2+6 y$
$27=y^2+6 y$
We have a quadratic equation for ' $y$ '. Solving for the roots of this equation we have, $y^2+6 y-27=0 y^2+9 y-3 y-27$ $=0 y(y+9)-3(y+9)=0(y+9)(y-3)=0$ The roots of the above equation are ' -9 ' and ' 3 '. Thus the ordinates of the other end of the line segment could be -9 or 3.
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Question 255 Marks
Prove that (2, -2) (-2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.
Answer
Let A(2, -2), B(-2, 1) and C(5, 2) be the given points.$\text{AB}=\sqrt{(-2-2)^2+(1+2)^2}$
$\Rightarrow\ \text{AB}=\sqrt{(-4)^2+(3)^2}$
$\Rightarrow\ \text{AB}=\sqrt{16+9}$
$\Rightarrow\ \text{AB}=\sqrt{25}$
$\text{BC}=\sqrt{(5+2)^2+(2-1)^2}$
$\Rightarrow\ \text{BC}=\sqrt{(7)^2+(1)^2}$
$\Rightarrow\ \text{BC}=\sqrt{49+1}$
$\Rightarrow\ \text{BC}=\sqrt{50}$
$\text{AC}=\sqrt{(5-2)^2+(2+2)^2}$
$\Rightarrow\ \text{AC}=\sqrt{(3)^2+(4)^2}$
$\Rightarrow\ \text{AC}=\sqrt{9+16}$
$\Rightarrow\ \text{AC}=\sqrt{25}$
$\text{AB}^2=(\sqrt{25})^2$
$\Rightarrow\ \text{AB}^2=25$
$\text{BC}^2=(\sqrt{50})^2$
$\Rightarrow\ \text{BC}^2=50$
Since, $AB^2 + AC^2 = BC^2$
$\therefore$ ABC is a right angled triangle.
Length of the hypotenuse $\text{BC}=\sqrt{50}=5\sqrt{2}$
$\text{Area of }\triangle\text{ABC}=\frac{1}{2}\times\text{AB}\times\text{AC}$
$=\frac{1}{2}\times\sqrt{25}\times\sqrt{25}$
$=\frac{25}{2}\text{square units}.$
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Question 265 Marks
If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices.
Answer
Two opposite points of a square are (5, 4) and (1, -6). Let ABCD be a square and A(5, 4) and C(1, -6) are the opposite points. Let the co-ordinates of B be (x, y). Join AC.

Now, $\because A B=B C$ (Sides of a square) $\Rightarrow A B^2=B C^2 \Rightarrow(x-5)^2+(y-4)^2=(x-1)^2+(y+6)^2 \Rightarrow x^2-10 x+25+y^2-8 y+$ $16=x^2-2 x+1+y^2+12 y+36 \Rightarrow-10 x+2 x-8 y-12 y=37-41 \Rightarrow-8 x-20 y=-4 \Rightarrow 2 x+5 y=1$ (Dividing by -4$) \Rightarrow$ $2 x=1-5 y \Rightarrow x=\frac{1-5 y}{2} \because A B C$ is a right angled triangle.
$\therefore A C^2=A B^2+B C^2 \Rightarrow(5-1)^2+(4+6)^2=x^2-10 x+25+y^2-8 y+16+x^2-2 x+1+y^2+12 y+36 \Rightarrow(4)^2+(10)^2=$ $2 x^2+2 y^2-12 x+4 y+78 \Rightarrow 16+100=2 x^2+2 y^2-12 x+4 y+78 \Rightarrow 2 x^2+2 y^2-12 x+4 y+78-16-100=0 \Rightarrow x^2+$ $y ^2-6 x +2 y -19=0$ (Dividing by 2) Substituting $x =\frac{1-5 y }{2}\left(\frac{1-5 y }{2}\right)^2+ y ^2-6\left(\frac{1-5 y }{2}\right)+2 y -19=0$ $\Rightarrow \frac{1+25 y^2-10 y}{4}+ y ^2-3(1-5 y)+2 y -19=0 \Rightarrow 1+25 y ^2-10 y +4 y ^2-12(1-15 y )+8 y -76=0 \Rightarrow 1+25 y ^2-$ $10 y+4 y^2-12+60 y+8 y-76=0 \Rightarrow 29 y^2+58 y-87=0 \Rightarrow y^2+2 y-3=0$ (Dividing by 29 ) $\Rightarrow y^2+3 y-y-3=0 \Rightarrow$ $y(y+3)-1(y+3)=0 \Rightarrow(y+3)(y-1)=0$ Either $y+3=0$, then $y=-3$ or $y-1=0$, then $y=1$ If $y=1$, then $x =\frac{1-5 y }{2}=\frac{1-5 \times 1}{2}$
$=\frac{1-5}{2}=\frac{-4}{2}=-$
If $y =-3$, then $x =\frac{1-5(-3)}{2}=\frac{1+15}{2}$ $=\frac{16}{2}=8$
$\therefore$ Other points are $(-2,1)$ and $(8,-3)$.
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Question 275 Marks
Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.
Answer
In ∆ABC, D and E are the mid points of the sides AB and AC respectively,

Then $\text{DE}=\frac{1}{2}\text{BC}$
Let the co-ordinates of the vertices of a $\triangle A B C$ be $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$.
Then coordinates of D will be,
$\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)$
Then coordinates of E will be,
$\Big(\frac{\text{x}_1+\text{x}_3}{2},\frac{\text{y}_1+\text{y}_3}{2}\Big)$
Now length of $\text{BC}=\sqrt{(\text{x}_2-\text{x}_3)^2+(\text{y}_2-\text{y}_3)^2}\ ....(\text{i})$
and length of $\text{DE}=\sqrt{\Big(\frac{\text{x}_1+\text{x}_2}{2}-\frac{\text{x}_1+\text{x}_3}{2}\Big)^2+\Big(\frac{\text{y}_1+\text{y}_2}{2}-\frac{\text{y}_1+\text{y}_3}{2}\Big)^2}$
$=\sqrt{\Big(\frac{\text{x}_1+\text{x}_2-\text{x}_1-\text{x}_3}{2}\Big)^2+\Big(\frac{\text{y}_1+\text{y}_2-\text{y}_1-\text{y}_3}{2}\Big)^2}$
$=\sqrt{\Big(\frac{\text{x}_2-\text{x}_3}{2}\Big)^2+\Big(\frac{\text{y}_2-\text{y}_3}{2}\Big)^2}$
$=\frac{1}{2}\sqrt{(\text{x}_2-\text{x}_3)^2+(\text{y}_2-\text{y}_3)^{2}}$
$=\frac{1}{2}\text{BC}$ [From (i)]
Hence proved.
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Question 285 Marks
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3).
Answer
The distance d between two points $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$ is given by the formula, $d =\sqrt{\left( x _1- x _2\right)^2+\left( y _1- y _2\right)^2}$ The centre of a circle is at equal distance from all the points on its circumference. Here it is given that the circle passes through the points $A(6,-6), B(3,-7)$ and $C(3,3)$. Let the centre of the circle be represented by the point $O(x, y)$. So we have $A O=B O=C O A O=\sqrt{(6-x)^2+(-6-y)^2} BO =\sqrt{(3-x)^2+(-7-y)^2}$
$CO =\sqrt{(3- x )^2+(3- y )^2}$ Equating the first pair of these equations we have,
$A O=B O$
$\sqrt{(6-x)^2+(-6-y)^2}=\sqrt{(3-x)^2+(-7-y)^2}$ Squaring on both sides of the equation we have, $(6-x)^2+$ $(-6-y)^2=(3-x)^2+(-7-y)^2 36+x^2-12 x+36+y^2+12 y=9+x^2-6 x+49+y^2+14 y 6 x+2 y=143 x+y=7$
Equating another pair of the equations we have, $AO = CO \sqrt{(6- x )^2+(-6- y )^2}=\sqrt{(3- x )^2+(3- y )^2}$
Squaring on both sides of the equation we have, $(6-x)^2+(-6-y)^2=(3-x)^2+(3-y)^2 36+x^2-12 x+36+y^2+12 y$ $=9+x^2-6 x+9+y^2-6 y 6 x-18 y=54 x-3 y=9$ Now we have two equations for ' $x$ ' and ' $y$ ', which are $3 x+y=7 x-$ $3 y=9$ From the second equation we have $y=-3 x+7$. Substituting this value of ' $y$ ' in the first equation we have, $x$ -
$3(-3 x+7)=9 x+9 x-21=910 x=30$
$x=3$
Therefore the value of ' $y$ ' is,
$y=-3 x+7$
$y=-3(3)+7$
$y=-2$
Hence the co-ordinates of the centre of the circle are $(3,-2)$.
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Question 295 Marks
In what ratio does the point (-4, 6) divide the line segment joining the points A(-6, 10) and B(3, -8)?
Answer
Let (-4, 6) Divide AB internally in the ratio k : 1 using three section formula, we get $(-4,6)=\Big(\frac{3\text{k}-6}{\text{k}+1},\frac{-8\text{k}+10}{\text{k}+1}\Big)$ So, $-4=\frac{3\text{k}-6}{\text{k}+1}$ i.e., -4k - 4 = 3k - 6i.e., 7k = 2
i.e., k : 1 = 2 : 7
You check for the y-coordinate also, So, the point (-4, 6) divides the line segment joining the points A(-6, 10) and B(3, -8) in the ratio 2 : 7.
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Question 305 Marks
If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the,
  1. interior,
  2. exterior of the triangle.
Answer
Let the third vertex of an equilateral triangle be $(x, y)$. Let $A(-4,3), B(4,3)$ and $C(x, y)$. We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal.. $\therefore A B=B C=C A$ $\Rightarrow A B^2=B C^2=C A^2$.......(i) Now taking first two parts.
$A B^2=B C^2$
$\Rightarrow(4+4)^2+(3-3)^2=(x-4)^2+(y-3)^2 \Rightarrow 64+0=x^2+16-8 x+y^2+9-6 y \Rightarrow x^2+y^2-8 x-6 y=39 \ldots . .$. (ii)Now, taking first and third parts.
$AB ^2= CA ^2$
$\Rightarrow(4+4)^2+(3-3)^2=(-4-x)^2+(3-y)^2 \Rightarrow 64+0=16+x^2+8 x+9+y^2-6 y \Rightarrow x^2+y^2+8 x-6 y=39$ subtracting eq.(ii) from eq.(iii), we get

⇒ x = 0Now, put the value of x in eq.(ii), we get
$0 + y^2 - 0 - 6y = 39$
$\Rightarrow y^2 - 6y - 39 = 0$
$\therefore\ \text{y}=\frac{6\pm\sqrt{(-6)^2-4(1)(-39)}}{2\times1}$
$\bigg[\because$ Solution of ax + bx + c = 0 is $\text{x}={-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}\over2\text{a}}\bigg]$
$\Rightarrow\ \text{y}=\frac{6\pm\sqrt{36+156}}{2}$
$\Rightarrow\ \text{y}=\frac{6\pm\sqrt{192}}{2}$
$\Rightarrow\ \text{y}=\frac{6\pm2\sqrt{48}}{2}$
$\Rightarrow\ \text{y}=3\pm\sqrt{48}$
$\Rightarrow\ \text{y}=3\pm4\sqrt{3}$ $\Rightarrow\ \text{y}=3+4\sqrt{3}$ or $3-4\sqrt{3}$ So, the points of third vertex are, $(0, 3+4\sqrt{3})$ or $(3-4\sqrt{3})$
But given that, the origin lies in the interior of the ∆ABC and the x-coordinate of third vertex is zero.
Then, y-coordinate of third vertex should be negative.
Hence, the require coordinate of third vertex,
$\text{C}=(0,3-4\sqrt{3}).[\text{C}\neq(0,3+4\sqrt{3})]$
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Question 315 Marks
If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of its centroid.
Answer
In $\triangle A B C, D, E$ and $F$ are the mid-points of the sides $B C, C A$ and $A B$ respectively. The co-ordinates of $D$ are $(-2,3)$, of $E$ are $(4,-3)$ and of $F$ are $(4,5)$. Let the co-ordinates of $A, B$ and $C$ be $\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right)$ respectively.

$\because D$ is mid point of $B C \therefore x=\frac{x_2+x_3}{2}$ and $y=\frac{y_2+y_3}{2} \Rightarrow-2=\frac{x_2+x_3}{2}$
$\Rightarrow x _2+ x _3=-4$
and $3=\frac{ y _2+ y _3}{2} \Rightarrow y _2+ y _3=6$
Similarly E and F are the mid points of AC and AB respectively. $\therefore 4=\frac{ x _3+ x _1}{2} \Rightarrow x _3+ x _1=8$
and $-3=\frac{ y _3+ y _1}{2} \Rightarrow y _3+ y _1=-6$
$4=\frac{ x _1+ x _2}{2} \Rightarrow x _1+ x _2=8$
and $5=\frac{ y _1+ y _2}{2} \Rightarrow y _1+ y _2=10$
Now, $x_2+x_3=-4$ $\qquad$ (i) $x_3+x_1=8$
......(ii) $x_1+x_2=8$ $\qquad$ (iii) Adding we get, $2\left( x _1+ x _2+ x _3\right)=12$
$\Rightarrow x _1+ x _2+ x _3=\frac{12}{2}$
$\Rightarrow x _1+ x _2+ x _3=6$ $\qquad$ (iv) Subtracting (i), (ii) and (iii) from
(iv) $x_1=10 x_2=-2 x_3=-2$ Similarly, $y_2+y_3=6$ $\qquad$ (v) $y _3+$
$y_1=-6$ $\qquad$ (vi) $y_1+y_2=10$ $\qquad$ (vii) Adding, we get $2\left( y _1+ y _2+ y _3\right)=10 \Rightarrow y _1+ y _2+ y _3=\frac{10}{2}$
$\Rightarrow y_1+y_2+y_3=5$ $\qquad$ (viii) Subracting (v), (vi) and (vii) from (viii) $y _1=-1 y _2=11 y _3=-5 \therefore$ Co-ordinates of $A , B$ and C will be $(10,-1),(-2,11),(-2,-5)=\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$ or $\left(\frac{10-2-2}{3}, \frac{-1+11-5}{3}\right)$ or $\left(\frac{6}{3}, \frac{5}{3}\right)$ or $\left(2, \frac{5}{2}\right)$
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Question 325 Marks
Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.
Answer

Let OBCD be the quadrilateral P, Q, R, S be the midpoint off OB, CD, OD and BC.
Let the coordinates of O, B, C, D are (0, 0), (x, 0), (x, y) and (0, y).
coordinates of P are $\Big(\frac{\text{x}}{2},0\Big)$
coordinates of Q are $\Big(\frac{\text{x}}{2},\text{y}\Big)$
coordinates of R are $\Big(0,\frac{\text{y}}{2}\Big)$
coordinates of S are $\Big(\text{x},\frac{\text{y}}{2}\Big)$
Coordinates of midpoint of PQ are $\bigg[\frac{\frac{\text{x}}{2}+\frac{\text{x}}{2}}{2},\frac{0+\text{y}}{2}\bigg]=\Big(\frac{\text{x}}{2},\frac{\text{y}}{2}\Big)$
Coordinates of midpoint of RS are $\bigg[\frac{(0+\text{x})}{2},\frac{\frac{\text{y}}{2}+\frac{\text{y}}{2}}{2}\bigg]=\Big[\frac{\text{x}}{2},\frac{\text{y}}{2}\Big]$
Since, the coordinates of the mid-point of PQ = coordinates of mid-point of RS
$\therefore$ PQ and RS bisect each other.
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Question 335 Marks
If $\text{D}\Big(\frac{-1}{2},\frac{5}{2}\Big),$ E(7, 3) and $\text{F}\Big(\frac{7}{2},\frac{7}{2}\Big)$ are the mid-points of sides of ∆ABC, find the area of ∆ABC.
Answer

The midpoint of BC is $\text{D}\Big(\frac{-1}{2},\frac{5}{2}\Big)$
The midpoint of AB is $\text{F}\Big(\frac{7}{2},\frac{7}{2}\Big)$
The midpoint of AC is E(7, 3),
Consider the line segment BC,
$\Rightarrow\ \frac{\text{p}+\text{r}}{2}=-\frac{1}{2};\ \frac{\text{q}+\text{s}}{2}=\frac{5}{2}$
$\Rightarrow\ \text{p}+\text{r}=-1;\ \text{q}+\text{s}=5\ .....(\text{i})$
Consider the line segment AB,
$\Rightarrow\ \frac{\text{p}+\text{x}}{2}=\frac{7}{2};\ \frac{\text{q}+\text{y}}{2}=\frac{7}{2}$
$\Rightarrow\ \text{p}+\text{x}=7;\ \text{q}+\text{y}=7\ .....(\text{ii})$
Consider the line segment AC,
$\Rightarrow\ \frac{\text{r}+\text{x}}{2}=7;\ \frac{\text{s}+\text{y}}{2}=3$
$\Rightarrow\ \text{r}+\text{x}=14;\ \text{s}+\text{y}=6\ ....(\text{iii})$
Solve (i), (ii) and (iii) to get A(x, y) = A(11, 4), B(p, q) = B(-4, 3), C(r, s) = C(3, 2)
Let us assume that BC is base of the triangle,
$\overline{\text{BC}}=\sqrt{(-4-3)^2+(3-2)^2}=\sqrt{50}$
Equation of the line BC is,
$\frac{\text{x}+4}{-4-3}=\frac{\text{y}-3}{3-2}$
$\Rightarrow\ \text{x}+7\text{y}-17=0$
The perpendicular distance from a point $P(x_1, y_1)$ is,
$\text{P}=\Big|\frac{1(11)+7(4)-17}{\sqrt{50}}\Big|=\frac{22}{\sqrt{50}}$
The area of the triangle is $\text{A}=\frac{1}{2}\times\sqrt{50}\times\frac{22}{\sqrt{50}}$
= 11sq. units
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Question 345 Marks
In Fig., a right triangle BOA is given C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.
Answer
We have a right angle triangle $\triangle\text{BOA,}$ right angled at O. Co-ordinates are B(0, 2b), A(2a, 0) and C(0, 0). We have to prove that mid-point C of hypotenuse AB is equidistant from the vertices. In general to find the mid-point P(x, y) of two points $A(x_1, y_1) and B(x_2, y_2)$ we use section formula as, $\text{P}(\text{x, y})=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)$So co-ordinates of C is,
$C ( a , b )$ In general, the distance between $A \left( x _1, y _1\right)$ and $B \left( x _2, y _2\right)$ is given by,
$\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
So, $\text{CO}=\sqrt{(\text{a}-0)^2+(\text{b}-0)^2}$ $=\sqrt{\text{a}^2+\text{b}^2}$ $\text{CB}=\sqrt{(\text{a}-0)^2+(\text{b}-2\text{b})^2}$ $=\sqrt{\text{a}^2+\text{b}^2}$ $\text{CA}=\sqrt{(\text{a}-2\text{a})^2+(\text{b}-0)^2}$ $=\sqrt{\text{a}^2+\text{b}^2}$Hence, mid-point C of hypotenuse AB is equidistant from the vertices.
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Question 355 Marks
Determine the ratio in which the point (-6, a) divides the join of A(-3, 1) and B(-8, 9). Also find the value of a.
Answer
The co-ordinates of a point which divided two points $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio m : n is given by the formula,$(\text{x},\text{y})=\bigg[\Big(\frac{\text{mx}_2+\text{nx}_1}{\text{m}+\text{n}}\Big),\Big(\frac{\text{my}_2+\text{ny}_1}{\text{m}+\text{n}}\Big)\bigg]$
Here we are given that the point P(-6, a) divides the line joining the points A(-3, 1) and B(-8, 9) in some ratio.
Let us substitute these values in the earlier mentioned formula.
$(-6,\text{a})=\bigg[\Big(\frac{\text{m}(-8)+\text{n}(-3)}{\text{m}+\text{n}}\Big),\Big(\frac{\text{m}(9)+\text{n}(1)}{\text{m}+\text{n}}\Big)\bigg]$Equating the individual components we have,
$-6=\frac{\text{m}(-8)+\text{n}(-3)}{\text{m}+\text{n}}$ $-6\text{m}-6\text{n}=-8\text{m}-3\text{n}$ $2\text{m}=3\text{n}$ $\frac{\text{m}}{\text{n}}=\frac{3}{2}$ We see that the ratio in which the given point divides the line segment is 3 : 2. Let us now use this ratio to find out the value of ‘a’. $(-6,\text{a})=\bigg[\Big(\frac{\text{m}(-8)+\text{n}(-3)}{\text{m}+\text{n}}\Big),\Big(\frac{\text{m}(9)+\text{n}(1)}{\text{m}+\text{n}}\Big)\bigg]$ $(-6,\text{a})=\bigg[\Big(\frac{3(-8)+2(-3)}{3+2}\Big),\Big(\frac{3(9)+2(1)}{3+2}\Big)\bigg]$ Equating the individual components we have $\text{a}=\frac{3(9)+2(1)}{3+2}$ $\text{a}=\frac{29}{5}$ Thus the value of 'a' is $\frac{29}{5}.$
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Question 365 Marks
In what ratio is the line segment joining the points (-2, -3) and (3, 7) divided by the y-axis? Also, find the coordinates of the point of division.
Answer
Let y-axis divides PQ in the ratio.$\lambda:1$
Let R(x, y) be the coordinates of the point of division.
Then, the coordinates of the point of division are,$\text{R}\bigg(\frac{3\times\lambda+(-2)\times1}{\lambda+1},\frac{7\times\lambda+(-3)\times1}{\lambda+1}\bigg)=\bigg(\frac{3\lambda-2}{\lambda+1},\frac{7\lambda-3}{\lambda+1}\bigg)$
Since R lies on y-axis and x-coordinates of every point on y-axis is zero.$\therefore\ \frac{3\lambda-2}{\lambda+1}=0$
$\Rightarrow\ 3\lambda-2=0$
$\Rightarrow\ 3\lambda=2$
$\Rightarrow\ \lambda=\frac{2}{3}$
Hence, the required ratio is $\frac{2}{3}:1$ i.e., 2 : 3 putting $\lambda=\frac{2}{3}$ in the coordinates of R, we get$\Rightarrow\ \frac{7\lambda-3}{\lambda+1}=\frac{7\times\frac{2}{3}-3}{\frac{2}{3}+1}$
$\Rightarrow\ \frac{\frac{14-9}{3}}{\frac{2+3}{3}}=\frac{\frac{5}{3}}{\frac{5}{3}}=1$
Hence, the coordinates of R(0, 1).
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Question 375 Marks
If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also, find the distances QR and PR.
Answer
Given Q(0, 1) is equidistant from P(-5, -3) and R(x, 6) so PQ = QR $\sqrt{(5-0)^2+(-3-1)^2}=\sqrt{(0-\text{x})^2+(1-6)^2}$ $\sqrt{(5)^2+(-4)^2}=\sqrt{(-\text{x})^2+(-5)^2}$ $\sqrt{25+16}=\sqrt{\text{x}^2+25}$$41=\text{x}^2+25$
$16=\text{x}^2$
$\text{x}=\pm4$
So, point R is (4, 6) or (-4, 6) When point R is (4, 6)$\text{PR}=\sqrt{(5-4)^2+(-3-6)^2}$
$=\sqrt{1^2+(-9)^2}=\sqrt{1+81}=\sqrt{82}$
$\text{QR}=\sqrt{(0-4)^2+(1-6)^2}$
$=\sqrt{(-4)^2+(-5)^2}=\sqrt{16+25}=\sqrt{41}$
When point R is (-4, 6)$\text{PR}=\sqrt{(5-(-4))^2+(-3-6)^2}$
$=\sqrt{(9)^2+(-9)^2}=\sqrt{81+81}=9\sqrt{2}$
$\text{QR}=\sqrt{(0-(-4))^2+(1-6)^2}$
$=\sqrt{(4)^2+(-5)^2}=\sqrt{16+25}=\sqrt{41}$
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Question 385 Marks
Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), $\text{B}(2+\sqrt{3},\ 5)$ and C(2, 6).
Answer
Three vertices of a ||gm ABCD are A(2, 4), $\text{B}(2+\sqrt{3},\ 5)$ and C(2, 6). Draw one diagonal AC of ||gm ABCD
$\because$ Diagonal bisects the ||gm into triangle equal in area
Area of $\triangle\text{ABC}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[2(5-6)+(2+\sqrt{3})(6-4)+2(4-5)]$
$=\frac{1}{2}[2\times(-1)+(2+\sqrt{3})\times2+2\times(-1)]$
$=\frac{1}{2}[-2+4+2\sqrt{3}-2]$
$=\frac{1}{2}(2\sqrt{3})=\sqrt{3}\text{ sq.units}$
$\therefore$ Area of ||gm ABCD = 2 × area $(\triangle\text{ABC})$
$=2\times(\sqrt{3})=2\sqrt{3}\text{ sq.units}$
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Question 395 Marks
Show that A(-3, 2), B(-5, -5), C(2, -3), and D(4, 4) are the vertices of a rhombus.
Answer
Vertices of a quadrilateral ABCD are A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4). Join the diagonals AC and BD which intersect each other at O.
Let O is the mid-point of AC, then co-ordinates of O will be $\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)$
or $\Big(\frac{-3+2}{2},\frac{2-3}{2}\Big)$ or $\Big(\frac{-1}{2},\frac{-1}{2}\Big)$ If O is the mid-point of BD, their co-ordinates of O will be. $\Big(\frac{-5+4}{2},\frac{-5+4}{2}\Big)$ or $\Big(\frac{-1}{2},\frac{-1}{2}\Big)$ $\because$ The co-ordinates of O in both case are same length of $\text{AC}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$ $=\sqrt{(-3-2)^2+(2+3)^2}$$=\sqrt{(-5)^2+(5)^2}=\sqrt{25+25}$
$=\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}$
and $\text{BD}=\sqrt{(-5-4)^2+(-5-4)^2}$ $=\sqrt{(-9)^2+(-9)^2}$ $=\sqrt{81+81}=\sqrt{162}=\sqrt{2\times81}=9\sqrt{2}$ Side $\text{AB}=\sqrt{(-3+5)^2+(2+5)^2}$ $=\sqrt{(2)^2+(7)^2}=\sqrt{4+49}=\sqrt{53}$ $\text{BC}=\sqrt{(2+5)^2+(-3+5)^2}$ $=\sqrt{(7)^2+(2)^2}=\sqrt{49+4}=\sqrt{53}$ $\text{CD}=\sqrt{(4-2)^2+(4+3)^2}$ $\text{CD}=\sqrt{(2)^2+(7)^2}$ $\text{CD}=\sqrt{4+49}$ $\text{CD}=\sqrt{53}$ and $\text{DA}=\sqrt{(-3-4)^2+(2-4)^2}$ $\text{DA}=\sqrt{(-7)^2+(-2)^2}$ $\text{DA}=\sqrt{49+4}$ $\text{DA}=\sqrt{53}$ $\because$ The diagonals bisect each other, but length of diagonals is not equal and sides are equal. $\therefore$ The figure is of a rhombus. Hence proved.
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Question 405 Marks
Which point on x-axis is equidistant from (5, 9) and (-4, 6)?
Answer
Let A(5, 9) and B(-4, 6) be the given points. Let C(x, 0) be the point on x-axis Now,$\text{AC}=\sqrt{(\text{x}-5)^2+(0-9)^2}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2+25-10\text{x}+(-9)^2}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2-10\text{x}+25+81}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2-10\text{x}+106}$
$\text{BC}=\sqrt{(\text{x}+4)^2+(0-6)^2}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+16+8\text{x}+(-6)^2}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+8\text{x}+16+36}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+8\text{x}+52}$
$Since, AC = BC Or, AC^2 = BC^2 x^2 - 10x + 106 = x^2 + 8x + 52$
$\Rightarrow -10x + 106 = 8x + 52 \Rightarrow -10x - 8x = 52 - 106$
⇒ -18x = -54 $\Rightarrow\ \text{x}=\frac{54}{18}$
⇒ x = 3 Hence the points on x-axis is (3, 0).
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Question 415 Marks
If P(9a - 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.
Answer
Let P(9a - 2, -b) divides AB internally in the ratio 3 : 1.
By section formula,
$9\text{a}-2=\frac{3(8\text{a})+1(3\text{a}+1)}{3+1}$
$\Bigg[\because$ internal section formula, the coordinates of point P divides the line segment joining the point $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ in the ratio $m_1 : m_2$ internally is $\bigg(\frac{\text{m}_2\text{x}_1+\text{m}_1\text{x}_2}{\text{m}_1+\text{m}_2},\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}\bigg)\Bigg]$
and $-\text{b}=\frac{3(5)+1(-3)}{3+1}$
$\Rightarrow9\text{a}-2=\frac{24\text{a}+3\text{a}+1}{4}$
and $-\text{b}=\frac{15-3}{4}$
$\Rightarrow9\text{a}-2=\frac{27\text{a}+1}{4}$
and $-\text{b}=\frac{12}{4}$
⇒ 36a - 8 = 27a + 1
and b = -3
⇒ 36a - 27a - 8 - 1 = 0
⇒ 9a - 9 = 0
a = 1
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Question 425 Marks
If G be the centroid of a triangle ABC and P be any other point in the plane, prove that $PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^2 + 3GP^2.$
Answer
Let $\triangle ABC$ be any triangle whose coordinates are $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$. Let $P$ be the origin and $G$ be the centraid of the triangle.
Now, we have to prove that
$PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^2 + 3GP^2 .......(i)$
We know that,
Coordinates of the centraid $G$ of the triangle $A B C$ whose coordinates are $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$ is given by,
$G\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$
The distance between $A \left( x _1, y _1\right)$ and $B \left( x _2, y _2\right)$ is given by,
$\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
So,
$PA^2 = (x_1 - 0)^2 + (y_1 - 0)^2$
$PA^2 = x_1^2 + y_1^2$
$PB^2 = (x_2 - 0)^2 + (y_2 - 0)^2$
$PB^2 = x_2^2 + y_2^2$
$PC^2 = (x_3 - 0)^2 + (y_3 - 0)^2$
$PC^2 = x_3^2 + y_3^2$
Now,
$\text{GP}^2=\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3}-0\Big)^2+\Big(\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}-0\Big)^2$
$=\frac{(\text{x}_1+\text{x}_2+\text{x}_3)^2}{9}+\frac{\text{(y}_1+\text{y}_2+\text{y}_3)^2}{9}$
$\text{GA}^2=\Big(\text{x}_1-\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3}\Big)^2+\Big(\text{y}_1-\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)^2$
$=\Big(\frac{\text{3x}_1-\text{x}_1-\text{x}_2-\text{x}_3}{3}\Big)^2+\Big(\frac{\text{3y}_1-\text{y}_1-\text{y}_2-\text{y}_3}{3}\Big)^2$
$=\frac{\text{(2x}_1-\text{x}_2-\text{x}_3)^2}{9}+\frac{\text(\text{2y}_1-\text{y}_2-\text{y}_3)^2}{9}$
$\text{GB}^2=\Big(\text{x}_2-\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3}\Big)^2+\Big(\text{y}_2-\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)^2$
$=\frac{(2\text{x}_2-\text{x}_1-\text{x}_3)^2}{9}+\frac{(2\text{y}_2-\text{y}_1-\text{y}_3)^2}{9}$
$\text{GC}^2=\Big(\text{x}_3-\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3}\Big)^2+\Big(\text{y}_3-\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)^2$
$\text{GC}^2=\frac{(3\text{x}_3-\text{x}_1-\text{x}_2)^2}{9}+\frac{(3\text{y}_3-\text{y}_1-\text{y}_2)^2}{9}$
Now, we get the value of left hand side of equation (i) as
$PA^2 + PB^2 + PC^2 = x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2$
Similarly, we get the value of right hand side of equation (i) as
$GA^2 + GB^2 + GC^2 + 3GP^2$​​​​​​​ $=\frac{(2\text{x}_1-\text{x}_2-\text{x}_3)^2}{9}+\frac{(2\text{y}_1-\text{y}_2-\text{y}_3)^2}{9}+\frac{(2\text{x}_2-\text{x}_1-\text{x}_3)^2}{9}+\frac{(2\text{y}_2-\text{y}_1-\text{y}_3)^2}{9}$$+\frac{(2\text{x}_3-\text{x}_1-\text{x}_2)^2}{9}+\frac{\text{(2y}_3-\text{y}_1-\text{y}_2)^2}{9}+3\times\Big[\frac{(\text{x}_1+\text{x}_2+\text{x}_3)^2}{9}+\frac{(\text{y}_1+\text{y}_2+\text{y}_3)^2}{9}\Big]$
$=\Big[\frac{2}{3}(\text{x}^2_1+\text{x}^2_2+\text{x}^2_3)+\frac{1}{3}(\text{x}^2_1+\text{x}^2_2+\text{x}^2_3)\Big]+\Big[\frac{2}{3}(\text{y}^2_1+\text{y}^2_2+\text{y}^2_3)+\frac{1}{3}(\text{y}^2_1+\text{y}^2_2+\text{y}^2_3)\Big]$
$= x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2$
$Hence, PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^2 + 3GP^2.$
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Question 435 Marks
If A(-3, 5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.
Answer
It is given that A(-3, 5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD.

Area of the quadrilateral ABCD $=\text{Area of }\triangle\text{ABC}+\text{Area of }\triangle\text{ACD}$
$\text{ar}(\triangle\text{ABC})=\frac{1}{2}|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)|$
$=\frac{1}{2}|-3(-7-(-8))+(-2)(-8-5)+1(5-(-7))|$
$=\frac{1}{2}|-3+26+12|$
$=\frac{35}{2}\text{ sq.units}$
$\text{ar}(\triangle\text{ACD})=\frac{1}{2}|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)|$
$=\frac{1}{2}|-3(-8-3)+1(3-5)+6(5-(-8))|$
$=\frac{1}{2}|33-2+78|$
$=\frac{109}{2}\text{ sq.units}$
$\therefore$ Area of the quadrilateral $\text{ABCD}=\frac{35}{2}+\frac{109}{2}=\frac{144}{2}=72\text{sq. units}$
Hence, the area of the given quadrilateral is 72 square units.
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Question 445 Marks
Prove that the points (2, 3), (-4, -6) and $\Big(1, \frac{3}{2}\Big)$ do not form a triangle.
Answer
Let the given points A(2, 3), B(-4, -6) and $\text{C}\Big(1, \frac{3}{2}\Big)$ Then, $\text{AB}=\sqrt{(-4-2)^2+(-6-3)^2}$ $=\sqrt{36+81}$ $=\sqrt{117}=\sqrt{9\times13}$ $=3\sqrt{13 }$ $\text{BC}=\sqrt{(1-(-4))^2+\Big(\frac{3}{2}-(-6)\Big)^2}$ $=\sqrt{25+\Big(\frac{3}{2}+6\Big)^2}$ $=\sqrt{25+\Big(\frac{15}{2}\Big)^2}$ $=\sqrt{25+\frac{225}{4}}$ $=\sqrt{\frac{100+225}{4}}$ $=\sqrt{\frac{335}{4}}$ $=\sqrt{83.75}$ $\text{AC}=\sqrt{(1-2)^2+\Big(\frac{3}{2}-3\Big)^2}$ $=\sqrt{1+\Big(\frac{3-6}{2}\Big)^2}$ $=\sqrt{1+\frac{9}{4}}=\sqrt{\frac{13}{4}}$$=\sqrt{3.25}$
We know that, In triangle sum of any two sides is greater than the third side. Since, $\text{AC} + \text{AB} \ngtr \text{BC}$Here, Sum of two sides is not greater than the third side.
Therefore, Points (2, 3), (-4, -6) and $\Big(1, \frac{3}{2}\Big)$ do not form a triangle.
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Question 455 Marks
Find the area of the quadrilaterals, the coordinates of whose vertices are (1, 2), (6, 2), (5, 3) and (3, 4).
Answer
Let the vertices of quadrilateral ABCD be A(1, 2), B(6, 2), C(5, 3) and D(3, 4). Join AC

We get two triangles: $\triangle\text{ABC}$ and $\triangle\text{ADC}$
Now area of $\triangle\text{ABC}$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[1(2-3)+6(3-2)+5(2-2)]$
$=\frac{1}{2}[1\times(-1)+6\times1+5\times0]$
$=\frac{1}{2}[-1+6+0]$
$=\frac{1}{2}(5)=\frac{5}{2}\text{ sq. units}$
and area of $\triangle\text{ADC}$
$=\frac{1}{2}[1(3-4)+5(4-2)+3(2-3)]$
$=\frac{1}{2}[1\times(-1)+5\times2+3\times(-1)]$
$=\frac{1}{2}[-1+10-3]$
$=\frac{1}{2}\times6=3\text{ sq. units}$
$\therefore$ Area of quadrilateral ABCD
$=\frac{5}{2}+3=\frac{5+6}{2}$
$=\frac{11}{2}\text{ sq. units}$
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Question 465 Marks
Find the area of the quadrilaterals, the coordinates of whose vertices are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Answer
Let the vertices of the quadrilateral be A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3).Join AC to form two triangle $\triangle\text{ABC}$ and $\triangle\text{ACD}.$

Area of quadrilateral ABCD $=|\text{Area of }\triangle\text{ABC}|+|\text{Area of }\triangle\text{ACD}|$
Area of $\triangle\text{ABC}$
$=\frac{1}{2}|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)|$
$=\frac{1}{2}|(-4)[(-5)-(-2)]+(-3)[(-2)-(-2)]+3[(-2)-(-5)]|$
$=\frac{1}{2}|(12+0+9)|=\frac{21}{2}\text{ sq. units}$
Area of $\triangle\text{ACD}$
$=\frac{1}{2}|(-4)[(-2)-(3)]+3[(3)-(-2)]+2((-2)-(-2))|$$=\frac{1}{2}|20+15+0|=\frac{35}{2}\text{ sq.units}$
Area of quadrilateral ABCD $=\text{Ar.}(\triangle\text{ABC})+\text{Ar.}(\triangle\text{ACD})$
$=\Big(\frac{21}{2}+\frac{35}{2}\Big)\text{ sq.units}$
$=28\text{ sq.units.}$
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Question 475 Marks
Find the ratio in which the point $\text{P}\Big(\frac{3}{4},\frac{5}{12}\Big)$ divides the line segment joining the points $\text{A}\Big(\frac{1}{2},\frac{3}{2}\Big)$ and B(2, -5).
Answer
Suppose $\text{P}\Big(\frac{3}{4},\frac{5}{12}\Big)$ divides the line segment joining the points $\text{A}\Big(\frac{1}{2},\frac{3}{2}\Big)$ and B(2, -5) in the ratio k : 1.Using section formula, we get
Coordinates of $\text{P}=\Bigg(\frac{2\text{k}+\frac{1}{2}}{\text{k}+1},\frac{-5\text{k}+\frac{3}{2}}{\text{k}+1}\Bigg)$
$\therefore\ \Bigg(\frac{2\text{k}+\frac{1}{2}}{\text{k}+1},\frac{-5\text{k}+\frac{3}{2}}{\text{k}+1}\Bigg)=\Big(\frac{3}{4},\frac{5}{12}\Big)$
$\Rightarrow\ \frac{2\text{k}+\frac{1}{2}}{\text{k}+1}=\frac{3}{4}$ and $\frac{-5\text{k}+\frac{3}{2}}{\text{k}+1}=\frac{5}{12}$
Now,
$\frac{2\text{k}+\frac{1}{2}}{\text{k}+1}=\frac{3}{4}$
⇒ 8k + 2 = 3k + 3
⇒ 5k = 1
$\Rightarrow\ \text{K}=\frac{1}{5 }$
Putting $\text{K}=\frac{1}{5 }$ in $\ \frac{-5\text{k}+\frac{3}{2}}{\text{k}+1}=\frac{5}{12},$ we get
$\text{LHS}=\frac{-5\times\frac{1}{5}+\frac{3}{2}}{\frac{1}{5}+1}=\frac{-1+\frac{3}{2}}{\frac{1}{5}+1}$
$=\frac{\frac{1}{2}}{\frac{6}{5}}=\frac{5}{12}=\text{RHS}$
Thus, the required ratio is $\frac{1}{5}:1$ or 1 : 5.
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Question 485 Marks
If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.
Answer
Let P(x, y), Q(2, 1), R(1, -2) be the given points,
$Here,x_1 = x, y_1 = y$
$x_2 = 2, y_2 = 1$
The distance between two points P(x, y) and Q(2, 1) is given by,
$\text{PQ}=\sqrt{(2-\text{x})^2+(1-\text{y})^2}$
Similarly,
$\text{PR}=\sqrt{(1-\text{x})^2+(-2-\text{y})^2}$
Now, both these distance are given to be the same
PQ = PR
$\sqrt{(2-\text{x})^2+(1-\text{y})^2}=\sqrt{(1-\text{x})^2+(-2-\text{y})^2}$
Squaring both the sides,
$\Rightarrow (2 - x)^2 + (1 - y)^2 = (1 - x)^2 + (-2 - y)^2$
$\Rightarrow 4 + x^2 -4x + 1 + y^2 - 2y = 1 + x^2 - 2x + 4 + y^2 + 4y$
$\Rightarrow 4 + x^2 - 4x + 1 + y^2 - 2y - 1 - x^2 + 2x - 4 - y^2 - 4y = 0$
$\Rightarrow -2x - 6y = 0$
$\Rightarrow -2(x + 3y) = 0$
$\Rightarrow x + 3y = 0$
Hence prove.
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Question 495 Marks
Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.
Answer
We have P(x, y), Q(-3, 0) and R(3, 0)
$\text{PQ}=\sqrt{(\text{x}+3)^2+(\text{y}-0)^2}$
$\Rightarrow\ 4=\sqrt{\text{x}^2+9+6\text{x}+\text{y}^2}$
Squaring both sides,
$\Rightarrow\ (4)^2=\Big(\sqrt{\text{x}^2+9+6\text{x}+\text{y}^2}\Big)^2$
$\Rightarrow\ 16=\text{x}^2+9+6\text{x}+\text{y}^2$
$\Rightarrow\ \text{x}^2+\text{y}^2=16-9-6\text{x}$
$\Rightarrow\ \text{x}^2+\text{y}^2=7-6\text{x}\ .....(1)$
$\text{PR}=\Big(\sqrt{(\text{x}-3)^2+(\text{y}-0)^2}\Big)$
$\Rightarrow\ 4=\sqrt{\text{x}^2+9-6\text{x}+\text{y}^2}$
Squaring both sides,
$\Rightarrow\ (4)^2=\Big(\sqrt{\text{x}^2+9-6\text{x}+\text{y}^2}\Big)^2$
$\Rightarrow\ 16=\text{x}^2+9-6\text{x}+\text{y}^2$
$\Rightarrow\ \text{x}^2+\text{y}^2=16-9+6\text{x}$
$\Rightarrow\ \text{x}^2+\text{y}^2=7+6\text{x}\ .....(2)$
Equating (1) and (2),
7 - 6x = 7 + 6x
⇒ 7 - 7 = 6x + 6x
⇒ 0 = 12x
⇒ x = 0
Substituting the value of x = 0 in (2)
$x^2 + y^2 = 7 + 6x$
$0 + y^2 = 7 + 6 \times 0$
$y^2 = 7$
$\text{y}=\pm\sqrt{7}$
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Question 505 Marks
Points P, Q, R and S divides the line segment joining A(1, 2) and B(6, 7) in 5 equal parts. Find the coordinates of the points P, Q and R.
Answer
Points P, Q, R and S divides AB in 5 equal parts and let coordinates of P, Q, R and S be
P divides AB in the ratio 1 : 4$\therefore\ \text{x}_1=\frac{1\times6+4\times1}{1+4}=\frac{6+4}{5}$
$=\frac{10}{5}=2$
$\Big\{\text{x}=\frac{\text{m}_1\text{x}_2+\text{m}_2\text{x}_1}{\text{m}_1+\text{m}_2}\Big\}$
$\text{y}_1=\frac{1\times7+4\times2}{1+4}=\frac{7+8}{5}$
$=\frac{15}{5}=3$
$\therefore$ Coordinates of P are (2, 3) Q divides AB in the ratio 2 : 3
$\therefore\ \text{x}_2=\frac{2\times6+3\times1}{2+3}=\frac{12+3}{5}$
$=\frac{15}{5}=3$
$\text{y}_2=\frac{2\times7+3\times2}{2+3}=\frac{14+6}{5}$
$=\frac{20}{5}=4$
$\therefore$ Coordinates of Q are (3, 4) R divides AB in the ratio 3 : 2
$\therefore\ \text{x}_3=\frac{3\times6+2\times1}{3+2}=\frac{18+2}{5}$
$=\frac{20}{5}=4$
$\text{y}_3=\frac{3\times7+2\times2}{3+2}=\frac{21+4}{5}$
$=\frac{25}{5}=5$
Coordinates of R are (4, 5).
S divides line AB in 4 : 1 ratio $\text{x}_4=\frac{4\times6+1\times1}{4+1}=\frac{24+1}{5}=\frac{25}{5}=5$ $\text{y}_4=\frac{4\times7+1\times2}{4+1}=\frac{28+2}{5}=\frac{30}{5}=6$ Therefore co-ordinate of $S(x_4, y_4) = (5, 6).$
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5 Marks Questions - MATHS STD 10 Questions - Vidyadip