2 Marks Questions - Maths STD 10 Questions - Vidyadip

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17 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = $\frac{3}{7}$ AB and P lies on the line segment AB.

Answer

A = (-2, -2) and B=(2, -4)
It is given that AP= $\frac{3}{7}$ AB
PB = AB - AP = AB −$\frac{3}{7}$AB = $\frac{4}{7}$AB
So, we have AP:PB = 3:4
Let coordinates of P be (x, y)
Using Section formula to find coordinates of P, we get
$x = \frac{{( - 2) \times 4 + 2 \times 3}}{{3 + 4}} = \frac{{6 - 8}}{7} = \frac{{ - 2}}{7}$
$y = \frac{{( - 2) \times 4 + ( - 4) \times 3}}{{3 + 4}} = \frac{{ - 8 - 12}}{7} = \frac{{ - 20}}{7}$
Therefore, Coordinates of point P are $\left( {\frac{{ - 2}}{7},\frac{{ - 20}}{7}} \right)$.

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Question 22 Marks

Find the coordinates of a point $A$, where $AB$ is the diameter of a circle whose centre is $(2, -3)$ and $B$ is $(1, 4).$

Answer

We want to find coordinates of point A. $AB$ is the diameter and coordinates of centerare $(2, -3)$ and, coordinates of point B are $(1, 4).$
Let coordinates of point A are (x, y). Using section formula, we get
$2 = \frac{{x + 1}}{2}$
$\Rightarrow 4 = x + 1$
$\Rightarrow x = 3$
Using section formula, we get
$ - 3 = \frac{{4 + y}}{2}$
$\Rightarrow 4 + y = −6$
$\Rightarrow y =− 6 − 4 = −10$
Therefore, Coordinates of point A are$ (3, −10).$

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Question 32 Marks

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer

Let A → (1, 2), B → (4, y), C→ (x, 6) and D→ (3, 5).
We know that the diagonals of parallelogram bisect each other.
So, Coordinates of the mid-point of diagonal AC
= Coordinates of the mid-point of diagonal BD
$\Rightarrow \left( {\frac{{1 + x}}{2},\frac{{2 + 6}}{2}} \right) = \left( {\frac{{4 + 3}}{2},\frac{{y + 5}}{2}} \right)$
$\Rightarrow \left( {\frac{{1 + x}}{2},4} \right) = \left( {\frac{7}{2},\frac{{y + 5}}{2}} \right)$
$\Rightarrow \frac{{1 + x}}{2} = \frac{7}{2}$
$\Rightarrow$ 1 + x = 7
$\Rightarrow$ x = 6
and $4 = \frac{{y + 5}}{2}$
$\Rightarrow$ y + 5 = 8
$\Rightarrow$ y = 3

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Question 42 Marks

Find the ratio in which the segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

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Question 52 Marks

Find the area of the rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
[Hint: Area of a rhombus $=\frac{1}{2}$ (product of its diagonals)]

Answer

Let A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1)
$AC = \sqrt {{{( - 1 - 3)}^2} + {{(4 - 0)}^2}} = 4\sqrt 2 $
$BD = \sqrt {{{( - 2 - 4)}^2} + {{( - 1 - 5)}^2}} = \sqrt {36 + 36} = 6\sqrt 2 $
Area of rhombus = $\frac{1}{2}{d_1} \times {d_2}$
$ = \frac{1}{2}AC \times BD$
$ = \frac{1}{2} \times 4\sqrt 2 \times 6\sqrt 2 = 24$ Sq. unit.

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Question 62 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

$PQ = RQ$
$\Rightarrow PQ^2 = RQ^2$
$\Rightarrow (0 - 5)^2 + [1 - (-3)]^2 = (0 - x)^2 + (1 - 6)^2$
$\Rightarrow 25 + 16 = x^2 + 25$
$\Rightarrow x^2 = 16$
$\Rightarrow x = \pm4$
$\therefore$ R $\rightarrow$ ($\pm$4, 6)
$QR = \sqrt {{{(0 \pm 4)}^2} + {{(1 - 6)}^2}} = \sqrt {41}$
$PR = \sqrt {{{( \pm 4 - 5)}^2} + {{\{ 6 - ( - 3)\} }^2}}$
$= \sqrt {{{(4 - 5)}^2} + 81} \;{\text{or}}\;\sqrt {{{( - 4 - 5)}^2} + 81}$
$= \sqrt {82} \;{\text{or}}\;9\sqrt 2$.

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Question 72 Marks

Find the values of y for which the distance between the points $P (2, -3)$ and $Q (10, y)$ is $10$ units.

Answer

Using Distance formula, we have
$10 = \sqrt {{{(2 - 10)}^2} + {{( - 3 - y)}^2}} $
$ \Rightarrow 10 = \sqrt {{{( - 8)}^2} + 9 + {y^2} + 6y} $ $$
$ \Rightarrow 10 = \sqrt {64 + 9 + {y^2} + 6y} $
Squaring both sides, we get
$100 = 73 + y^2 + 6y$
$\Rightarrow y^2 + 6y - 27 = 0$
Solving this Quadratic equation by factorization, we can write
$\Rightarrow y^2 + 9y − 3y - 27 = 0$
$\Rightarrow y(y + 9) - 3(y + 9) = 0$
$\Rightarrow (y + 9)(y − 3) = 0$
$\Rightarrow y = 3, −9$

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Question 82 Marks

Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

Answer

Let A $\rightarrow$ (5, -2), B $\rightarrow$ (6, 4) and C $\rightarrow$ (7, -2)
Then,
$AB = \sqrt {{{(6 - 5)}^2} + {{(4 - ( - 2))}^2}} = \sqrt {{{(1)}^2} + {{(6)}^2}}$
$ = \sqrt {1 + 36} = \sqrt {37}$
$BC = \sqrt {{{(7 - 6)}^2} + {{( - 2 - 4)}^2}} = \sqrt {{{(1)}^2} + {{( - 6)}^2}}$
$= \sqrt {1 + 36} = \sqrt {37}$
CA = $\sqrt {{{(7 - 5)}^2} + {{( - 2 - (-2))}^2}} = \sqrt {{{(2)}^2} + {{(0)}^2}}$
We see that AB = BC $\neq$ CA
So, the A, B and C are vertices of an isosceles triangle.

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Question 92 Marks

Determine if the points $(1, 5), (2, 3)$ and $(-2, -11)$ are collinear.

Answer

The points of trisection means that the points which divide the line into three equal parts. From the figure, it is clear that C, and D are these two points. Let $C (x_1, y_1)$ and $D (x_2, y_2)$ are the points of trisection of the line segment joining the given points i.e., $BC = CD = DA$
Let $BC = CD = DA = k$, Point C divides BC and CA as: $BC = kCA = CD + DA = k + k = 2k$
Hence the ratio between BC and CA is: $\frac{B C}{C A}=\frac{k}{2 k}=\frac{1}{2}$
Therefore, point C divides BA internally in the ratio 1:2 then by section formula we have that if a point P(x, y) divides two points $P (x_1, y_1)$ and $Q (x_2, y_2)$ in the ratio m:n then, the point (x, y) is given by $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}\right)$
Therefore C(x, y) divides B(–2, –3) and A(4,–1) in the ratio 1:2, then
$C(x, y)=\left(\frac{(1 \times 4)+(2 \times-2)}{1+2}, \frac{(1 \times-1)+(2 \times-3)}{1+2}\right)$
$C(x, y)=\left(\frac{4-4}{1+2}, \frac{-1-6}{1+2}\right)$
$C(x, y)=\left(0, \frac{-7}{3}\right)$
Point D divides the BD and DA as:$DA = kBD = BC + CD = k + k = 2k$
Hence the ratio between BD and DA is: $\frac{B D}{D A}=\frac{2 k}{k}=\frac{2}{1}$
The point D divides the line BA in the ratio 2:1
So now applying section formula again we get,
$D(x, y)=\left(\frac{(2 \times 4)+(1 \times-2)}{2+1}, \frac{(2 \times-1)+(1 \times-3)}{2+1}\right)$
$D(x, y)=\left(\frac{8-2}{3}, \frac{-2-3}{3}\right)$
$D(x, y)=\left(\frac{6}{3}, \frac{-5}{3}\right)$
$D(x, y)=\left(2, \frac{-5}{3}\right)$

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Question 102 Marks

Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Answer

Distance between two points
Given: Points A(0, 0), B(36, 15)

For two points $A \left( x _1, y _1\right)$ and $B \left( x _2, y _2\right)$,
Distance is given by $f = \sqrt{{({\mathrm{{x_2-x_1}}})^2}+{({\mathrm{{y_2-y_1}}})^2}}$Distance between (0, 0) and (36, 15) is:
$f = \sqrt{{({\mathrm{{36-0}}})^2}+{({\mathrm{{15-0}}})^2}}$
$f = \sqrt{1296+225}$
$f = \sqrt{1521}$= 39
Hence Distance between points A and B is 39 units
Yes, we can find the distance between the given towns A and B. Let us take town A at origin point (0, 0)
Hence, town B will be at point (36, 15) with respect to town A
And, as calculated above, the distance between town A and B will be 39 km

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Question 112 Marks

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).

Answer

It is given that P (x,y) is equidistant from A(3, 6) and B(-3, 4).
Using Distance formula, we can write
PA = PB
$\sqrt {{{(x - 3)}^2} + {{(y - 6)}^2}} = \sqrt {{{[x - ( - 3)]}^2} + {{(y - 4)}^2}} $
$ \Rightarrow \sqrt {{x^2} + 9 - 6x + {y^2} + 36 - 12y} = \sqrt {{x^2} + 9 + 6x + {y^2} + 16 - 8y} $
Squaring both sides, we get
$ \Rightarrow $$ x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 - 8y$
$\Rightarrow −6x − 12y + 45 = 6x - 8y + 25$
$\Rightarrow 12x + 4y = 20$
$\Rightarrow 3x + y = 5$

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Question 122 Marks

Find the coordinates of the point of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(–7, 4).

Answer

Let P and Q be the points of trisection of AB i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are
$\left(\frac{1(-7)+2(2)}{1+2}, \frac{1(4)+2(-2)}{1+2}\right), \text { i.e., }(-1,0)$
Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
$\left(\frac{2(-7)+1(2)}{2+1}, \frac{2(4)+1(-2)}{2+1}\right), \text { i.e., }(-4,2)$
Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).

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Question 132 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 the section formula, we get
$( - 4,6 ) = \left( \frac { 3 k - 6 } { k + 1 } , \frac { - 8 k + 10 } { k + 1 } \right)$
So, $- 4 = \frac { 3 k - 6 } { k + 1 }$
i.e., -4k - 4 = 3k - 6
i.e., 7k = 2
i.e., k:1 = 2:7
The same can be checked for the y-coordinate also.
Therefore, the ratio in which the point (-4,6) divides the line segment AB is 2: 7.

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Question 142 Marks

Find the coordinates of the point which divides the line segment joining the points (4, -3) and (8, 5) in the ratio 3 : 1 internally.

Answer

Let coordinates of the required point be R(x, y). This means R divides the join of P(4, -3) and Q(8, 5) in the ratio 3:1 internally.
Using the Section formula for internal division, here $x_1= 4, y_1= -3, x_2= 8, y_2= 5, m = 3, n = 1$
$\Rightarrow$(x,y) = $\left( \frac { mx _ { 2 } +n x _ { 1 } } { m+n} , \frac { m y _ { 2 } + ny _ { 1 } } { m+n } \right)$
$\Rightarrow$(x,y) = ($\frac { 3 ( 8 ) + 1 ( 4 ) } { 3 + 1 }$,$\frac { 3 ( 5 ) + 1 ( - 3 ) } { 3 + 1 }$)
$\Rightarrow$ (x,y) = $( \frac { 24 + 4 } { 4 } , \frac { 15-3} { 4 })$
$\Rightarrow$ (x,y) = $( \frac { 28 } { 4 } , \frac { 12} { 4 }) =(7, 3)$
$\Rightarrow$ x= 7 and y= 3
Thus, the coordinates of R (x,y) = (7,3)

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Question 152 Marks

Find a point on the y-axis which is equidistant from the points $A(6, 5)$ and $B (- 4, 3).$

Answer

We have to find a point on the y-axis which is equidistant from the points $A(6, 5)$ and $B (- 4, 3)$.We know that a point on y-axis is of the form $(0, y)$. So, let the required point be $P (0, y).$
Then,
$PA = PB$
$\Rightarrow \sqrt { ( 0 - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \sqrt { ( 0 + 4 ) ^ { 2 } + ( y - 3 ) ^ { 2 } }$
$\Rightarrow$ $36 + (y - 5)^2 = 16 + (y - 3)^2$
$\Rightarrow$ $36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9$
$\Rightarrow$ $4y = 36$
$\Rightarrow$ $y = 9$
So, the required point is (0, 9).

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Question 162 Marks

Find the relation between x and y such that the point $(x, y)$ is equidistant from the points $(7, 1)$ and $(3, 5).$

Answer

Let $P(x, y)$ be equidistant from the points $A(7, 1)$ and $B(3, 5)$
$AP = BP$ (Given)
$ \Rightarrow AP^2 = BP^2$
$ \Rightarrow (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$
$ \Rightarrow x^2 + 49 - 14x + y^2 + 1 - 2y = x^2 + 9 - 6x + y^2 + 25 - 10y$
$ \Rightarrow 49 - 14x ^ + 1 - 2y = 9 - 6x + 25 - 10y$
$ \Rightarrow - 14x + 6x - 2y + 10 y = 34 - 50$
$ \Rightarrow -8x + 8y = -16$
$ \Rightarrow x - y = 2$

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Question 172 Marks

Fig. given below shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give a reason for your answer.

Answer

Using the distance formula, we have
$A B=\sqrt{(6-3)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$
$\mathrm{BC}=\sqrt{(8-6)^{2}+(6-4)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$
$A C=\sqrt{(8-3)^{2}+(6-1)^{2}}=\sqrt{25+25}=\sqrt{50}=5 \sqrt{2}$
Since, $\mathrm{AB}+\mathrm{BC}=3 \sqrt{2}+2 \sqrt{2}=5 \sqrt{2}=\mathrm{AC}$
So, we can say that points A, B and C are collinear. Therefore, they are seated in a line.

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2 Marks Questions - Maths STD 10 Questions - Vidyadip