3 Marks Question - MATHS STD 10 Questions - Vidyadip

Questions

3 Marks Question

🎯

Test yourself on this topic

35 questions · timed · auto-graded

Question 13 Marks

Find the coordinate of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5).

Answer

Let the required points be P(x, y), then
PA = PB = PC.
The points A, B, C are (5,3), (5, -5) and (1, -5) and (1, -5) recpectively
$\Rightarrow PA^2 = PB^2= PC^2$
$\Rightarrow PA^2 = PB^2 and PB^2 = PC^2$
$PA^2 = PB^2$
$\Rightarrow (5 - x)^2+ (3 - y)^2 = (5 - x)^2 + (-5 - y)^2$
$25 + x^2- 10x + 9 + y^2- 6y = 25 + x^2- 10x + 25 + y^2 + 10y$
$-6y - 10y = 25 - 9 \Rightarrow -16y = 16$
$y = -1$
and $PB^2 = PC^2$
$\Rightarrow (5 - x)^2 + (-5 - y)^2 = (1 - x) + (-5 - y)^2$
$25 + x^2 - 10x + 25 + y^2 + 10y = 1 + x^2 - 2x + 25 + y^2 + 10y$
$-10x + 2x = -24 \Rightarrow -8x = -24$
$\text{x}=\frac{-24}{-8}=3$
Hence, the point P is (3, -1)

View full question & answer→

Question 23 Marks

Show that the following points are the vertices of a rectangle:
A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3)

Answer

Let A(-4, -1), B(-2, -4) C(4, 0) and D(2, 3) are the vertices of quadrilateral ABCD.
Then,

$\text{AB}=\sqrt{(-2+4)^2+(-4+1)^2}$
$=\sqrt{(2)^2+(-3)^2}$
$=\sqrt{4+9}=\sqrt{15}\text{ units}$
$\text{BC}=\sqrt{(4+2)^2+(0-4)^2}$
$=\sqrt{(6)^2+(-4)^2}$
$=\sqrt{36+16}=\sqrt{52}=7\sqrt{3}\text{ units}$
$\text{DC}=\sqrt{(2-4)^2+(3-0)^2}$
$=\sqrt{(-2)^2+(3)^2}$
$=\sqrt{4+9}=\sqrt{15}\text{ units}$
$\text{AD}=\sqrt{(-4-2)^2+(-1-3)^2}$
$=\sqrt{(-6)^2+(-4)^2}$
$=\sqrt{36+16}=\sqrt{52}=7\sqrt{3}\text{ units}$
Thus, $\text{AB}=\text{DC}$ and $\text{AD}=\text{BC}$
$\text{Diag}.\text{AC}=\sqrt{(4+4)^2+(0+1)^2}$
$=\sqrt{(8)^2+(1)^2}$
$=\sqrt{64+1}$
$=\sqrt{65}\text{ units}$
$\text{Diag}.\text{BD}=\sqrt{(2+2)^2+(3+4)^2}$
$=\sqrt{(4)^2+(7)^2}$
$=\sqrt{16+49}$
$=\sqrt{65}\text{ units}$
$\therefore\text{Diag}.\text{AC}\not=\text{Diag}.\text{BD}$
Thus, ABCD is a quadrilateral whose opposite are equal and the diagonals are equal.
Hence, quad. ABCD is a rectangle.

View full question & answer→

Question 33 Marks

Show taht the following points are collinear:A(2, -2), B(-3, 8) and C(-1, 4)

Answer

Let $A(x_1 = 2, y_1= -2), B(x_2 = -3, y_2= 8)$ and $C(x_3 = -1, y_3= 4)$ be the given points.
Now
$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$
$= 2(8 - 4) + (-3)(4 + 2) + (-1)( -2 -8)$
$= 8 - 18 + 10$
$= 0$
Hence the given point are collinear.

View full question & answer→

Question 43 Marks

If three consecutive vertices of a parallelogram ABCD are A(1, -2), B(3, 6) and C(5, 10), find its fourth vertex D.

Answer

Let A(1, -2), B(3, 6) and C(5, 10) are the given vertices of the parallelogram ABCD.

Let D(a, b) be its fourth vertex. Join AC and BD. Let AC and BD intersect at the point O. We know that the diagonal of a parallelogram bisect each other. So, O is the mid-point AC as well as that of BD. Mid-point of AC is $\Big(\frac{1+5}{2},\frac{-2+10}{2}\Big)\text{i.e},(3,4)$ Mid-point of BD is $\Big(\frac{3+\text{a}}{2},\frac{6+\text{b}}{2}\Big)$ $\frac{3+\text{a}}{2}=3$ and $\frac{6+\text{b}}{2}=4$ ⇒ a = 3 and b = 2 Hence the fourth vertices is D(3, 2).

View full question & answer→

Question 53 Marks

ABCD is a rectangle formed by points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). If P, Q, R and S be the midpoints of AB, BC, CD and DA respectively, show that PQRS is a rhombus.

Answer

ABCD is a rectangle and P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
Thus, we have
Coordinates of P $=\Big(\frac{-1-1}{2},\frac{{-1+4}}{2}\Big)=\Big(-1,\frac{3}{2}\Big)$
Coordinates of Q $=\Big(\frac{-1+5}{2},\frac{4+4}{2}\Big)=(2,4)$
Coordinates of R $=\Big(\frac{5+5}{2},\frac{-1+4}{2}\Big)=\Big(5,\frac{3}{2}\Big)$
Coordinates of S $=\Big(\frac{-1+5}{2},\frac{-1-1}{2}\Big)=(2,-1)$
Now,
Lenght of PQ $=\sqrt{(-2-1)^2+\Big(4-\frac{3}{2}\Big)^2}$
$=\sqrt{(-3)^2+\Big(\frac{5}{2}\Big)^2}$
$=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}$
Lenght of QR $=\sqrt{(5-2)^2+\Big(\frac{3}{2}-4\Big)^2}$
$=\sqrt{(3)^2+\Big(\frac{-5}{2}\Big)^2}$
$=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}$
Lenght of RS $=\sqrt{(2-5)^2+\Big(-1-\frac{3}{2}\Big)^2}$
$=\sqrt{(3)^2+\Big(\frac{-5}{2}\Big)^2}$
$=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}$
Lenght of SP $=\sqrt{(-1-2)^2+\Big(\frac{3}{2}+1\Big)^2}$
$=\sqrt{(3)^2+\Big(\frac{-5}{2}\Big)^2}$
$=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}$
Lenght of PR $=\sqrt{(5+1)^2+\Big(\frac{3}{2}-\frac{3}{2}\Big)^2}$
$=\sqrt{(6)^2+0}=\sqrt{36}=6$
Lenght of SQ $=\sqrt{(2-2)^2+(4+1)^2}$
$=\sqrt{0+(5)^2}=\sqrt{25}=5.$
Here, all sides of PQRS are equal but diagonals are not equal.
Hence, PQRS is a rhombus.

View full question & answer→

Question 63 Marks

If the points P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS, find the value of a and b

Answer

Let P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS.

Join the diagonal PR and SQ. Then intersect each other at the point O. We know that the diagonal of a parallelpgram bisect each other. Now, mid-point of PR is $\Big(\frac{\text{a}+2}{2},\frac{-11+15}{2}\Big)\text{i.e},\Big(\frac{\text{a}+2}{2},2\Big)$ And mid-point of SQ is $\Big(\frac{5+1}{2},\frac{\text{b}+1}{2}\Big)\text{i.e},\Big(3,\frac{\text{b}+1}{2}\Big)$ $\therefore\ \frac{\text{a}+2}{2}=3$ and $\frac{\text{b}+1}{2}=2$ ⇒ a = 4 and b = 3 Hence the required values are a = 4 and b = 3.

View full question & answer→

Question 73 Marks

Show that the following points are the vertices of a rectangle:
A(2, -2), B(14, 10), C(11, 13) and D(-1, 1)

Answer

Let A(2, -2), B(14, 10), C(11, 13) and D(-1, 1)be the angular points of quad. ABCD. Then,

$\text{AB}=\sqrt{(14-2)^2+(10+2)^2}$ $=\sqrt{(12)^2+(12)^2}$ $=\sqrt{144+144}=\sqrt{288}=12\sqrt{2}\text{ units}$ $\text{BC}=\sqrt{(11-14)^2+(13-10)^2}$ $=\sqrt{(-3)^2+(3)^2}$ $=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\text{ units}$ $\text{DC}=\sqrt{(-1-11)^2+(1-13)^2}$ $=\sqrt{(-12)^2+(-12)^2}$ $=\sqrt{144+144}=\sqrt{288}=12\sqrt{2}\text{ units}$ $\text{AD}=\sqrt{(-1-2)^2+(1+2)^2}$ $=\sqrt{(-3)^2+(3)^2}$ $=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\text{ units}$ Thus, $\text{AB}=\text{DC}$ and $\text{AD}=\text{BC}$ $\text{Diag}.\text{AC}=\sqrt{(11-2)^2+(13+2)^2}$ $=\sqrt{(9)^2+(15)^2}$ $=\sqrt{81+150}$ $=\sqrt{306}=3\sqrt{34}\text{ units}$ $\text{Diag}.\text{BD}=\sqrt{(-1-14)^2+(1-10)^2}$ $=\sqrt{(-15)^2+(-9)^2}$ $=\sqrt{225+81}$ $=\sqrt{306}=3\sqrt{34}\text{ units}$ $=\sqrt{65}\text{ units}$ $\therefore\text{Diag}.\text{AC}=\text{Diag}.\text{BD}$ Thus, ABCD is a quadrilateral whose opposite sides are equal and diagonals are equal. Hence, quad. ABCD is a rectangle.

View full question & answer→

Question 83 Marks

For what value of k(k > 0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to 53 square units?

Answer

Let $A(x_1 = -2, y_1= 5), B(x_2 = k, y_2= -4)$ and $C(x_3 = 2k + 1, y_3= 10)$ be the vertices of the triangle, so
$\text{ar}\big(\triangle\text{ABC)}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\Rightarrow53=\frac{1}{2}[(-2)(-4-10)+\text{k}(10-5)+(2\text{k}+1)(5+4)]$
$\Rightarrow53=\frac{1}{2}[28+5\text{k}+9(2\text{k}+1)]$
$\Rightarrow28+5\text{k}+18\text{k}+9=106$
$\Rightarrow37+23\text{k}=106$
$\Rightarrow23\text{k}=106-37=69$
$\Rightarrow\text{k}=\frac{69}{23}=3$
Hence, k = 3

View full question & answer→

Question 93 Marks

Find the coordinates of the midpoint the line segment joining:

A(3, 0) and B(-5, 4)
P(-11, -8) and Q(8, -2).

Answer

The coordinates of mid-points of the line segment joining A(3, 0) and B(-5, 4) are $\Big(\frac{3-5}{2},\frac{0+4}{2}\Big)$ or (-1, 2)
Let M(x, y) be the mid-point of AB, where A is (-11, -8) and B is (8, -2). Then,

$\text{x}=\frac{8-11}{2}=\frac{-3}{2}$ and $\text{y}=\frac{-8-2}{2}=\frac{-10}{2}=-5$

Hence, the required point is $\Big(\frac{-3}{2},-5\Big)$

View full question & answer→

Question 103 Marks

Find the area of $\triangle\text{ABC}$ whose vertices are:
A(-5, 2), B(-4, -5) and C(4, 5)

Answer

A(-5, 7), B(-4, -5) and C(4, 5) are the vertices of $\triangle\text{ABC}.$
Then, $(x_1 = -5, y_1= 7), (x_2 = -4, y_2 = -5)$ and $(x_3 = 4, y_3 = 5)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{(-5)(-5-5)+(-4)+(5-7)\\+(-4-2)+(4)(7-(-5))\big\}$
$=\frac{1}{2}\big\{(-5)(-10)-4(-2)+4(12)\big\}$
$=\frac{1}{2}\big\{50+8+48\big\}$
$=\frac{1}{2}(106)$
$=53\ \text{sq}.\text{units}$

View full question & answer→

Question 113 Marks

Find the distance between the points:
P(a + b, a - b) and Q(a - b, a + b)

Answer

The given points are P(a + b, a - b) and Q(a - b, a + b)
Then,$ [x_1 = (a + b), y_1 = (a - b)]$ and $[x_2 = (a - b)$, and $y_2 = (a + b)]$
$\therefore\text{PQ}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{a}-\text{b}-\text{a}-\text{b})^2+(\text{a}+\text{b}-\text{a}+\text{b})^2}$
$=\sqrt{(-\text{2b})^2+(\text{2b})^2}$
$=\sqrt{\text{4b}^2+\text{4b}^2}$
$=\sqrt{\text{8b}^2}=2\sqrt{2\text{b}}\text{ units}.$

View full question & answer→

Question 123 Marks

Show that the points A(3, 0), B(6, 4) and C(-1, 3) are the vertices of an isosceles right triangle.

Answer

Given: A(3, 0), B(6, 4) and C(-1, 3)
Now,
$\text{AB}=\sqrt{(6-3)^2+(4-0)^2}=\sqrt{(3)^2+(4)^2}$
$=\sqrt{9+16}=\sqrt{25}=5\text{ units}$
$\text{BC}=\sqrt{(-1-6)^2+(3-4)^2}=\sqrt{(-7)^2+(-1)^2}$
$=\sqrt{49+1}=\sqrt{50}=\sqrt{50}\text{ units}$
$\text{AC}=\sqrt{(-1-3)^2+(3-0)^2}=\sqrt{(-4)^2+(3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
We find that AB = AC = 5 units
$\therefore\triangle\text{ABC}$ is an isosceles treangle.
And,
$=\text{AB}^2+\text{AC}^2$
$=5^2+5^2$
$=25+25=50$
$\text{BC}^2=(\sqrt{50})^2=50$
$\Rightarrow\text{AB}^2+\text{AC}^2=\text{BC}^2$
This shows that $\triangle\text{ABC}$ is righta-angled at B.
Hence, the geven points are the vertices of an isosceles right triangle.

View full question & answer→

Question 133 Marks

If the point C(k, 4) divedes the join of A(2, 6) and B(5, 1) in the ratio 2 : 3 then find the value of k.

Answer

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3. So
$\text{k}=\frac{2\times5+3\times2}{2+3}$
$=\dfrac{10+6}{5}$
$=\frac{16}{5}$
Hence, $\text{k}=\frac{16}{5}.$

View full question & answer→

Question 143 Marks

Find the ratio in which the point P(m, 6) divides the join of A(-4, 3) and B(2, 8). Also find the value of m.

Answer

Let P dividing the join of line segment A(-4, 3) and B(2, 8) in the ratio k : 1.
$\therefore$ The point P is
$\Big(\frac{\text{k}\times2+1\times(-4)}{\text{k}+1},\frac{\text{k}\times8+1\times3}{\text{k}+1}\Big)=\Big(\frac{2\text{k}-4}{\text{k}+1},\frac{8\text{k}+3}{\text{k}+1}\Big)$
$\Rightarrow\ \frac{2\text{k}-4}{\text{k}+1}=\text{m}\dots(1)$
And
$\frac{8\text{k}+3}{\text{k}+1}=6$
Or $8\text{k}+3=6\text{k}+6$ or $2\text{k}=3\ \therefore\ \text{k}=\frac{3}{2}$
Putting value of k in (1)
$\frac{2\times\frac{3}{2}-4}{\frac{3}{2}+1}=\text{m}$
$\text{m}=\frac{3-4}{\frac{5}{2}}$
$\text{m}=-\frac{2}{5}$
Hence, $\text{m}=-\frac{2}{5},\text{k}=\frac{3}{2}$

View full question & answer→

Question 153 Marks

The midpoint P of the line segment joining the points A(-10, 4) and B(-2, 0) lies on the line segment joining the points C(-9, -4) and D(-4, y). Find the ratio in which P divides CD. Also find the value of y.

Answer

Since P is the mid-point of line segment joining the points A(-10, 4) and B(-2, 0).
Coordinates of P $=\Big(\frac{-10-2}{2},\frac{4+0}{2}\Big)=(-6,2)$
Let P(-6, 2) divides the line segment joining the points C(-9, -4) and D(-4, y) in the ratio k : 1.
$\therefore$ By section formula, we have
Coordinates of P
$=\Big(\frac{\text{k}\times(-4)+1\times(-9)}{\text{k}+1},\frac{\text{k}\times\text{y}+1\times(-4)}{\text{k}+1}\Big)=\Big(\frac{-4\text{k}-9}{\text{k}+1},\frac{\text{ky}-4}{\text{k}+1}\Big)$
But, coordinates of P are (-4, 2).
$\therefore\ \frac{-4\text{k}-9}{\text{k}+1}=-6\Rightarrow\ -4\text{k}-9=-6\text{k}-6$
$\Rightarrow\ 2\text{k}=3\Rightarrow\ \text{k}=\frac{3}{2}$
Thus, required ratio is 2 : 3.
Also, $\frac{\text{ky}-4}{\text{k}+1}=2$
$\Rightarrow\ \text{ky}-4=2\text{k}+2$
$\Rightarrow\ \frac{3}{2}\text{y}-4=2\times\frac{3}{2}+2$
$\Rightarrow\ \frac{3}{2}\text{y}=9$
$\Rightarrow\ \text{y}=6$

View full question & answer→

Question 163 Marks

The line segment joining the points A(3, -4) and B(1, 2) is trisected at the points P(p, -2) and $\text{Q}\Big(\frac{5}{3},\text{q}\Big).$ Find the values of p and q.

Answer

Point P divides the join of A(3, -4) and B(1, 2) in the ratio 1 : 2
Coordinates of P are:
$\Big(\frac{1\times1+2\times3}{1+2},\frac{1\times2+2\times(-4)}{1+2}\Big)$ or $\Big(\frac{7}{3},\frac{-6}{3}\Big)$ or $(\frac{7}{3}, -2)$
Also the point P is (p, -2) ⇒ $\text{P}=\frac{7}{3}$
Further Q is the midpoint of PB when
$\text{P}\Big(\frac{7}{3},-2\Big)$ and B(1, 2)
$\therefore$ coordinates of Q are $\bigg(\frac{\frac{7}{3}+1}{2},\frac{-2+2}{2}\bigg)$ or $\Big(\frac{5}{3},0\Big)$
Also, Q is $\Big(\frac{5}{3},\text{q}\Big)\Rightarrow\text{q}=0$
Hence, $\text{p}=\frac{7}{3}$ and q = 0

View full question & answer→

Question 173 Marks

Point A lies on the line segment PQ joinning P(6, -6) and Q(-4, -1) in such a way that $\frac{\text{PA}}{\text{PQ}}=\frac{2}{5}.$ If the point A also lies on the line 3x + k(y + 1) = 0, find the value of k.

Answer

$\frac{\text{PA}}{\text{PQ}}=\frac{2}{5}$
$\Rightarrow\frac{\text{PQ}}{\text{PA}}=\frac{5}{2}$
$\Rightarrow\frac{\text{PA}+\text{AQ}}{\text{PA}}=\frac{5}{2}$
$\Rightarrow1+\frac{\text{AQ}}{\text{PA}}=\frac{5}{2}$
$\Rightarrow\frac{\text{AQ}}{\text{PA}}=\frac{7}{2}-1=\frac{5-2}{2}=\frac{3}{2}$
$\frac{\text{PA}}{\text{AQ}}=\frac{2}{3}$
$\Rightarrow\text{PA}:\text{AQ}=2:3$
$\therefore\text{Coordinates of A}=\Big(\frac{2\times(-4)+3\times6}{2+3},\frac{2\times(-1)+3\times(-6)}{2+3}\Big)$
$=\Big(\frac{-8+18}{5},\frac{-2-18}{5}\Big)$
$=\Big(\frac{10}{5},\frac{-20}{5}\Big)$
$=(2,-4)$
Since the point A(2, -4) lies on the line 3
× k(y + 1), we have 3 × 2 + k(-4 + 1) = 0
⇒ 6 - 3k = 0
⇒ 3k - 6
⇒ k = 2

View full question & answer→

Question 183 Marks

Show that the following points are the vertices of a square:
A(3, 2), B(0, 5), C(-3, 2) and D(0, -1)

Answer

The angular points of quadrilateral ABCD are A(3, 2), B(0, 5), C(-3, 2) and D(0, 1)

$\text{AB}=\sqrt{(0-3)^2+(5-2)^2}$
$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
$\text{BC}=\sqrt{(-3-0)^2+(2-5)^2}$
$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
$\text{CD}=\sqrt{(0+3)^2+(-1-2)^2}$
$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
$\text{DA}=\sqrt{(3-0)^2+(2+1)^2}$
$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
$\therefore\text{AB}=\text{BC}=\text{CD}+\text{DA}=3\sqrt{2}$
$\text{Diag}.\text{AC}=\sqrt{(-3-3)^2+(2-2)^2}=6$
$\text{Diag}.\text{BD}=\sqrt{(0-0)^2+(-1-5)^2}=6$
$\therefore\text{Diag}.\text{AC}=\text{Diag}.\text{BD}$
Thus, all sides of quad. ABCD are equal and diagonals are also equal
Quad. ABCD is a square.

View full question & answer→

Question 193 Marks

In what ratio is the line segments joining the points A(-2, -3) and B(3, 7) divided by the y-axis? Also, find the coordinates of the point of division.

Answer

Let the y-axis cut the join A(-2, -3) and B(3, 7) at the point P in the ratio k : 1. Then,
By section formula, the co-ordinates of P are
$\text{P}\Big(\frac{3\text{k}-2}{\text{k}+1},\frac{7\text{k}-3}{\text{k}+1}\Big)$
But P lies on the y-axis so, its abscissa is 0.
$\therefore\ \frac{3\text{k}-2}{\text{k}+1}=0\Rightarrow\ 3\text{k}-2=0\Rightarrow\text{k}=\frac{2}{3}$
So the required ratio is $\frac{2}{3}:1$ which is 2 : 3
Putting $\text{k}=\frac{2}{3}$ in $\Big(0,\frac{7\text{k}-3}{\text{k}-1}\Big)$ we get the point P as
$\text{P}\bigg(0,\frac{7\times\frac{2}{3}-3}{\frac{2}{3}+1}\bigg)_{\text{i.e},\text{P}(0,1)}$
Hence the points of intersection of AB and the y-axis is P(0, 1) and P divides AB in the ratio 2 : 3.

View full question & answer→

Question 203 Marks

Find the value of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.

Answer

We have,
$PQ = 10$
$\Rightarrow PQ^2 = 100$
$\Rightarrow (9 - x)^2 + (10 - 4)^2 = 100$
$\Rightarrow (9 - x)^2 + 62 = 100$
$\Rightarrow (9 - x)^2 = 64$
$\Rightarrow 9-\text{x}=\pm8$
$\Rightarrow 9 - x = 8$ or $9 - x = -8$
$\Rightarrow x = 1 or x = 17$
Hence, the required values of x are 1 and 17.

View full question & answer→

Question 213 Marks

Show that the points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus.
Hint: Area of a rhombus $=\frac{1}{2}\times(\text{product of its diagonals}).$

Answer

Let A(-3, 2), B(-5, -5) C(2, -3) and D(4, 4) be the angular points of quad. ABCD. Join AC and BD.

Now,
$\text{AB}=\sqrt{(-5+3)^2+(-5-2)^2}$
$=\sqrt{(-2)^2+(-7)^2}$
$=\sqrt{4+49}=\sqrt{53}\text{ units}$
$\text{BC}=\sqrt{(2+5)^2+(-3+5)^2}$
$=\sqrt{(7)^2+(2)^2}$
$=\sqrt{4+49}=\sqrt{53}\text{ units}$
$\text{CD}=\sqrt{(4-2)^2+(4+3)^2}$
$=\sqrt{(2)^2+(7)^2}$
$=\sqrt{4+49}=\sqrt{53}\text{ units}$
$\text{DA}=\sqrt{(-3-4)^2+(2-4)^2}$
$=\sqrt{(-7)^2+(2)^2}$
$=\sqrt{4+49}=\sqrt{53}\text{ units}$
Thus, $\text{AB}=\text{BC}=\text{CD}=\text{DA}=\sqrt{53}\text{ units}$
$\text{Diag}.\text{AC}=\sqrt{(2-3)^2+(-3-2)^2}$
$=\sqrt{(5)^2+(-5)^2}$
$=\sqrt{(25)+(25)}$
$=\sqrt{50}=5\sqrt{2}\text{ units}$
$\text{Diag}.\text{BD}=\sqrt{(4+5)^2+(4+5)^2}$
$=\sqrt{(81)+(81)}$
$=\sqrt{162}=9\sqrt{2}\text{ units}$
$\therefore\text{Diag}.\text{AC}\not=\text{Diag}.\text{BD}$
Thus, ABCD is a quadrilateral having all sides equal but diagonals are unequal.
Hence, ABCD is a rhombus.
$\text{Area of rhombus ABCD}$ $=\frac{1}{2}\times\text{product of its diagonals}.$
$=\Big(\frac{1}{2}\times\text{AC}\times\text{BD}\Big)=\Big(\frac{1}{2}\times5\sqrt{2}\times9\sqrt{2}\Big)$
$=45\text{ sq.units}$

View full question & answer→

Question 223 Marks

Find the value of k so that area of the triangle with triangle with vertices A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.

Answer

Let $A(x_1, y_1) = A(k + 1, 1), B(x_2, y_2) = B(4, -3)$ and $C(x_3, y_3) = C(7, -k).$
Now,
$\text{ar}\big(\triangle\text{ABC)}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\Rightarrow6=\frac{1}{2}[(\text{k+1})(-3+\text{k})+4(-\text{k}-1)+7(1+3)]$
$\Rightarrow6=\frac{1}{2}[\text{k}^2-2\text{k}-3-4\text{k}-4-28]$
$\Rightarrow\text{k}^2-6\text{k}+9=0$
$\Rightarrow\text{(k}-3)^2=0$
$\Rightarrow\text{k}=3$
Hence, $\text{k}=3$

View full question & answer→

Question 233 Marks

Find the centroid of a $\triangle\text{ABC}$ whose vertices are A(-1, 0), B(5, -2) and C(8, 2)

Answer

Here $(x_1 = -1, y_1 = 0)(x_2 = 5, y_2 = -2)$ and $(x_3 = 8, y_3 = 2)$
Let G(x, y) be the centroid of $\triangle\text{ABC},$ then
$\text{x}=\frac{1}{3}(\text{x}_1+\text{x}_2+\text{x}_3)$
$=\frac{1}{3}(-1+8+5)=4$
$\text{y}=\frac{1}{3}(\text{y}_1+\text{y}_2+\text{y}_3)$
$=\frac{1}{3}(0-2+2)=0$
Hence, the centroid of $\triangle\text{ABC},$ is G(4, 0).

View full question & answer→

Question 243 Marks

Show that the points $\text{O}(0, 0),\text{A}(3,\sqrt{3})$ and $\text{B}(3,-\sqrt{3})$ are the vertices of an equilateral triangle. Find the area of this triangle.

Answer

Let O(0, 0), $\text{A}(3,\sqrt{3})$ and $\text{B}(3,-\sqrt{3})$ are the given points
$\therefore\text{OA}=\sqrt{(3-0)^2+(\sqrt{3}-0)^2}$
$=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}\text{ units}$
$\text{AB}=\sqrt{(3-3)^2+(-\sqrt{3}-\sqrt{3})^2}$
$=\sqrt{0^2+(-2\sqrt{3})}=\sqrt{12}=2\sqrt{3}\text{ units}$
$\text{OB}=\sqrt{(3-0)^2+(\sqrt{3}-0)^2}$
$=\sqrt{(3)^2+(-\sqrt{3})^2}$
$=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}\text{ units}$
$\therefore\text{OA}=\text{AB}=\text{AB}=2\sqrt{3}\text{ units}$
Hence, DABC is equilateral and each of its sides being $2\sqrt{3}\text{ units}$
Area of $\triangle\text{ABC}=\Big[\frac{\sqrt{3}}{4}\times(\text{side})^2\Big]\text{sq.unit}$
$=\Big[\frac{\sqrt{3}}{4}\times(2\sqrt{3})^2\Big]\text{sq.unit}$
$=\Big[\frac{\sqrt{3}}{4}\times4\times3\Big]\text{sq.unit}$
$=3\sqrt{3}\text{ sq.unit}$

View full question & answer→

Question 253 Marks

In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(-6, 9)?

Answer

Let P divided the join of A(8, 2), B(-6, 9) in the ratio k : 1. By section formula, the coordinates of p are.

$\Big(\frac{-6\text{k}+8}{\text{k}+1},\frac{9\text{k}+2}{\text{k}+1}\Big)$ $\therefore\ \frac{-6\text{k}+8}{\text{k}+1}=2$ and $\frac{9\text{k}+2}{\text{k}+1}=5$ $\Rightarrow\ -6\text{k}+8=2\text{k}+2$ and $9\text{k}+2=5\text{k}+5$ $\Rightarrow\ -8\text{k}=-6$ and $4\text{k}=3$ $\Rightarrow\ \text{k}=\frac{-6}{-8}=\frac{3}{4}$ and $\text{k}=\frac{3}{4}$ $\Rightarrow\ \text{k}=\frac{3}{4}$ in each case Hence, the required ratio of $\Big(\frac{3}{4}:1\Big)$ which is (3 : 4)

View full question & answer→

Question 263 Marks

If G(-2, 1) is the centroid of a $\triangle\text{ABC}$ and two of its vertices are A(1, -6) and B(-5, 2), find the third vertex of the triangle.

Answer

Two vertices of $\triangle\text{ABC}$ are A(1, -6) and B(-5, 2) let the third vertex be C(a, b). Then,
The co-ordinates of its centroid are
$\text{G}\Big(\frac{1-5+\text{a}}{3},\frac{-6+2+\text{b}}{3}\Big)\text{i.e}\ \text{G}\Big(\frac{-4+\text{a}}{3},\frac{-4+\text{b}}{3}\Big)$
But given that the centroid is G(-2, 1)
$\frac{-4+\text{a}}{3}=-2$ and $\frac{-4+\text{b}}{3}=1$
-4 + a = -6 and -4 + b = 3
a = -2 and b = 7
Hence, the third vertex C of $\triangle\text{ABC}$ is (-2, 7).

View full question & answer→

Question 273 Marks

If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.

Answer

Let the point P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
Then,
$PA = PB$
$\Rightarrow PA^2 = PB^2$
$\Rightarrow (x- 7)^2 + (y - 1)^2= (x - 3)^2 + (y - 5)^2$
$\Rightarrow x^2 + y^2 - 14x - 2y + 50 = x^2 + y^2 - 6x - 10y + 34$
$\Rightarrow 8x - 8y = 16$
$\Rightarrow x - y = 2$

View full question & answer→

Question 283 Marks

Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a Parallelogram. Show that ABCD is not a rectangle.

Answer

Let A(1, 2), B(4, 3) C(6, 6) and D(3, 5) be the angular points of a quadrilateral ABCD. Join AC and BD.
Now,
$\text{AB}=\sqrt{(4-1)^2+(3-2)^2}$
$=\sqrt{(3)^2+(1)^2}$
$=\sqrt{9+1}=\sqrt{10}\text{ units}$
$\text{BC}=\sqrt{(6-4)^2+(6-3)^2}$
$=\sqrt{(2)^2+(3)^2}$
$=\sqrt{4+9}=\sqrt{13}\text{ units}$
$\text{CD}=\sqrt{(3-6)^2+(6-3)^2}$
$=\sqrt{(-3)^2+(-1)^2}$
$=\sqrt{9+1}=\sqrt{10}\text{ units}$
$\text{DA}=\sqrt{(3-1)^2+(5-2)^2}$
$=\sqrt{(2)^2+(3)^2}$
$=\sqrt{4+9}=\sqrt{13}\text{ units}$
$\therefore\text{AB}=\text{CD}$ and $\text{BC}=\text{DA}$
Thus, ABCD is a parallelogram since its opposite sides are equal.
$\text{Diag}.\text{AC}=\sqrt{(6-1)^2+(6-2)^2}$
$=\sqrt{(5)^2+(4)^2}$
$=\sqrt{25+16}$
$=\sqrt{41}\text{ units}$
$\text{Diag}.\text{BD}=\sqrt{(3-4)^2+(5-3)^2}$
$=\sqrt{(-1)^2+(2)^2}$
$=\sqrt{1+4}$
$=\sqrt{5}\text{ units}$
$\therefore\text{Diag}.\text{AC}\not=\text{Diag}.\text{BD}$
Hence, it is not a rectangle.

View full question & answer→

Question 293 Marks

Points P, Q and R in that order are dividing a line segment joining A(1, 6) and B(5, -2) in four equal parts. Find the coordinates of P, Q and R.

Answer

Points P, Q, R divide the line segment joining the points A(1, 6) and (5, -2) in to four equal parts
⇒ Point P divide AB in the ratio 1 : 3 where A(1, 6), (5, -2)

Therefore, the point P is,
$\Big(\frac{1\times5+3\times1}{1+3},\frac{1\times(-2)+3\times6}{1+3}\Big)$ or $\Big(\frac{8}{4},\frac{16}{4}\Big)$ or $(2, 4)$
Now, Q is the midpoint of AB
The point Q is the midpoint of AB
The point Q is $\Big(\frac{1+5}{2},\frac{6-2}{2}\Big)$ or $(3,2)$
Also, R is the midpoint of the line segment joining Q(3, 2) and B(5, -2)
$\therefore$ The point R is $\Big(\frac{3+5}{2},\frac{2-2}{2}\Big)$ or $(4, 0)$

View full question & answer→

Question 303 Marks

If P(x, y) is a point equidistant from the points A(6, - 1) and B(2, 3), show that x - y = 3.

Answer

Let A(6, -1) and B(2, 3) be the given point and P(x, y) be the rwquired point, we get
$PA = PB$
$\Rightarrow (PA)^2 = (PB)^2$
$\Rightarrow (x - 6)^2 + (y + 1)^2 = (2 - x)^2 + (3 - y)^2$
$\Rightarrow 36 + x^2- 12x + y^2 + 1 + 2y = 4 + x^2 - 4x + 9 + y^2 - 6y$
$\Rightarrow -12x + 4x + 2y + 6y = 4 + 9 - 1 - 36$
$\Rightarrow -8x + 8y = -24$
$\Rightarrow -8(x - y) = -24$
$\Rightarrow x - y = 3$
Hence, x - y = 3

View full question & answer→

Question 313 Marks

Find the area of $\triangle\text{ABC}$ whose vertices are:
A(10, -6), B(2, 5) and C(-1, 3)

Answer

A(10, -6), B(2, 5) and C(-1, -3) are the vertices of $\triangle\text{ABC}.$
Then, $(x_1 = 10, y_1= -6), (x_2 = 2, y_2 = 5)$ and $(x_3 = -1, y_3 = 3)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{10(5-3)+2(3-(-6))+(-1)(-6-5)\big\}$
$=\frac{1}{2}\big\{10(2)+2(9)-1(-11)\big\}$
$=\frac{1}{2}\big\{20+18+11\big\}$
$=\frac{1}{2}(49)$
$=24.5\ \text{sq}.\text{units}$

View full question & answer→

Question 323 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

Let the required ratio be k : 1.
Then, by section formula,
Coordinates of P $=\bigg(\frac{\text{k}\times2+1\times\frac{1}{2}}{\text{k}+1},\frac{\text{k}\times(-5)+1\times\frac{3}{2}}{\text{k}+1}\bigg)$
$=\bigg(\frac{2\text{k}+\frac{1}{2}}{\text{k}+1},\frac{-5\text{k}+\frac{3}{2}}{\text{k}+1}\bigg)$
$=\bigg(\frac{4\text{k}+1}{2(\text{k}+1)},\frac{-10\text{k}+3}{2(\text{k}+1)}\bigg)$
Given, coordinates of P $=\Big(\frac{3}{4},\frac{5}{12}\Big)$
$\therefore\ \frac{4\text{k}+1}{2(\text{k}+1)}=\frac{3}{4}$
$\Rightarrow\ 16\text{k}+4=6\text{k}+6$
$\Rightarrow\ 10\text{k}=2$
$\Rightarrow\ \text{k}=\frac{1}{5}$
So, the required ratio is 1 : 5.

View full question & answer→

Question 333 Marks

Find the coordinate of the point which divides the join of A(-1, 7) and B(4, -3) in the ratio 2 : 3.

Answer

The point of AB are A(-1, 7) and B(4, -3)
$\therefore$ $(x_1 = -1, y_1 = 7)$ and $(x_2 = 4, y_2 = -3)$
Also m = 2 and n = 3

Let the required point be P(x, y)
By section formula, we have
$\Rightarrow\text{x}=\frac{(\text{mx}_2+\text{nx}_1)}{(\text{m+n})},\text{y}=\frac{(\text{my}_2+\text{ny}_1)}{(\text{m+n})}$
$\Rightarrow\text{x}=\frac{2\times4+3\times(-1)}{(2+3)},\text{y}=\frac{(2\times-3+3\times7)}{(2+3)}$
$\Rightarrow\text{x}=\frac{8-3}{5},\frac{-6+21}{5}$
$\Rightarrow\text{x}=\frac{5}{5},\frac{15}{5}$ or x = 1, y = 3
Hence the required point is P(1, 3)

View full question & answer→

Question 343 Marks

If the point P(k - 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.

Answer

If the point P(k - 1, 2), A(3, k) and B(k , 5).
$\because$ AP = BP
$\because$ $AP^2 = BP^2$
$\Rightarrow (k - 1 - 3)^2 + (2 - k)^2 = (k - 1 - k)^2 + (2 - 5)^2$
$\Rightarrow (k - 4)^2 + (2 - k)^2 = (-1)^2 + (-3)^2$
$\Rightarrow k^2 - 8y + 16 + 4 + k^2 - 4k = 1 + 9$
$\Rightarrow k^2 - 6y + 5 = 0$
$\Rightarrow (k - 1)(k - 5) = 0$
$\Rightarrow k = 1 or k = 5.$
Hence, k = 1 or k = 5.

View full question & answer→

Question 353 Marks

Show that the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.

Answer

Given: A(7, 10), B(-2, 5) and C(3, -4)
Now,
$\text{AB}=\sqrt{(-2-7)^2+(5-10)^2}=\sqrt{(-9)^2+(-5)^2}$
$=\sqrt{81+25}=\sqrt{106}=\sqrt{106}\text{ units}$
$\text{BC}=\sqrt{(3+2)^2+(-4-5)^2}=\sqrt{(5)^2+(-9)^2}$
$=\sqrt{25+18}=\sqrt{20}=\sqrt{106}\text{ units}$
$\text{AC}=\sqrt{(3+7)^2+(-4-10)^2}=\sqrt{(-4)^2+(-14)^2}$
$=\sqrt{16+196}=\sqrt{212}=\sqrt{212}\text{ units}$
We find that $\text{ AB} = \text{BC}=\sqrt{106}$
$\therefore\triangle\text{ABC}$ is an isosceles treangle.
And,
$\text{AB}^2+\text{BC}^2$
$=(\sqrt{106})^2+(\sqrt{106})^2$
$=106+106=212$
$\text{AC}^2=(\sqrt{212})^2=212$
$\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2$
This shows that $\triangle\text{ABC}$ is righta-angled at B.
Hence, the geven points are the vertices of an isosceles right triangle.

View full question & answer→

Generate Paper Free All Coordinate Geometry topics

3 Marks Question - MATHS STD 10 Questions - Vidyadip