ML Aggarwal Solution Class 10 Chapter 11 Section Formula MCQs

 MCQs

Choose the correct answer from the given four options (1 to 12) :

Question 1

The points A (9, 0), B (9, 6), C ( – 9, 6) and D ( – 9, 0) are the vertices of a

(a) rectangle

(b) square

(c) rhombus

(d) trapezium

Sol :

A (9, 0), B (9, 6), C (-9, 6), D (-9, 0)

$A B^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$

$=(9-9)^{2}+(6-0)^{2}=0^{2}+6^{2}$

$=0^{2}+36=36$

$C D^{2}=(-9+9)^{2}+(6-0)^{2}$

$=0^{2}+6^{2}=0+36=36$

$B C^{2}=(9+9)^{2}+(6-6)^{2}$

$=18^{2}+0^{2}=324+0=324$

$A D^{2}=(9+9)^{2}+0^{2}$

$=18^{2}+0^{2}=324+0=324$

AB=CD and BC=AD

But these are opposite sides of a rectangle

ABCD is a rectangle.

Ans (a)

Question 2

The mid-point of the line segment joining the points A ( – 2, 8) and B ( – 6, – 4) is

(a) ( – 4, – 6)

(b) (2, 6)

(c) ( – 4, 2)

(d) (4, 2)

Sol :

Mid-point of the line segment joining the points A (-2, 8), B (-6, -4)

$=\left(\frac{-2+6}{2}, \frac{8+4}{2}\right)$ or $\left(\frac{4}{2}, \frac{12}{2}\right)$

=(2,6)

Ans (b)

Question 3

If $P\left(\frac{a}{3}, 4\right)$ segment joining the points Q ( – 6, 5) and R ( – 2, 3), then the value of a is

(a) – 4

(b) – 6

(c) 12

(d) – 12

Sol :

$P\left(\frac{a}{3}, 4\right)$  is mid-point of the line segment

joining the points Q (-6, 5) and R (-2, 3)

$\therefore \frac{a}{3}=\frac{-6-2}{2}=\frac{-8}{2}=-4$

$a=-4 \times 3 \Rightarrow a=-12$

Ans (d)

Question 4

If the end points of a diameter of a circle are A ( – 2, 3) and B (4, – 5), then the coordinates of its centre are

(a) (2, – 2)

(b) (1, – 1)

(c) ( – 1, 1)

(d) ( – 2, 2)

Sol :

End points of a diameter of a circle are (-2, 3) and B (4,-5)

then co-ordinates of the centre of the circle

$=\left(\frac{-2+4}{2}, \frac{3-5}{2}\right)$ or $\left(\frac{2}{2}, \frac{-2}{2}\right)$

=(1,-1)

Ans (b)

Question 5

If one end of a diameter of a circle is (2, 3) and the centre is ( – 2, 5), then the other end is

(a) ( – 6, 7)

(b) (6, – 7)

(c) (0, 8)

(d) (0, 4)

Sol :

One end of a diameter of a circle is (2, 3) and centre is (-2, 5)

Let (x, y) be the other end of the diameter

$\frac{2+x}{2}=-2 \Rightarrow 2+x=-4$

$\Rightarrow x=-4-2=-6$

and $\frac{3+y}{2}=5 \Rightarrow 3+y=10$

$\Rightarrow y=10-3=7$

$\therefore$ Co-ordinates of other end are (-6,7)

Ans (a)

Question 6

If the mid-point of the line segment joining the points P (a, b – 2) and Q ( – 2, 4) is R (2, – 3), then the values of a and b are

(a) a = 4, b = – 5

(b) a = 6, b = 8

(c) a = 6, b = – 8

(d) a = – 6, b = 8

Sol :

the mid-point of the line segment joining the

points P (a, b – 2) and Q (-2, 4) is R (2, -3)

$2=\frac{a-2}{2} \Rightarrow a-2=4$

$\Rightarrow a=4+2=6$

$-3=\frac{b-2+4}{2}=\frac{b+2}{2}$

$\Rightarrow b+2=-6 \Rightarrow b=-6-2=-8$

$\therefore a=6, b=-8$

Ans (c)

Question 7

The point which lies on the perpendicular bisector of the line segment joining the points A ( – 2, – 5) and B (2, 5) is

(a) (0, 0)

(b) (0, 2)

(c) (2, 0)

(d) ( – 2, 0)

Sol :

the line segment joining the points A (-2, -5) and B (2, -5), has mid-point

$=\left(\frac{-2+2}{2}, \frac{-5+5}{2}\right)=(0,0)$

(0, 0) lies on the perpendicular bisector of AB. 

Ans (a)

Question 8

The coordinates of the point which is equidistant from the three vertices of ∆AOB (shown in the given figure) are

(a) (x, y)

(b) (y, x)

(c) $\left(\frac{x}{2}, \frac{y}{2}\right)$

(d) $\left(\frac{y}{2}, \frac{x}{2}\right)$

Sol :

In the given figure, vertices of a ∆OAB are (0, 0), (0, 2y) and (2x, 0)

The point which is equidistant from O, A and B is the mid-point of AB.

∴ Coordinates are $\left(\frac{0+2 x}{2}, \frac{2 y+0}{2}\right)$ or (x, y) (a)

Question 9

The fourth vertex D of a parallelogram ABCD whose three vertices are A ( – 2, 3), B (6, 7) and C (8, 3) is

(a) (0, 1)

(b) (0, – 1)

(c) ( – 1, 0)

(d) (1, 0)

Sol :

ABCD is a ||gm whose vertices A (-2, 3), B (6, 7) and C (8, 3).

The fourth vertex D will be the point on which diagonals AC and BD

bisect each other at O.

$\therefore$ Co-ordinates of $\mathrm{O}$ are $\left(\frac{-2+8}{2}, \frac{3+3}{2}\right)$ or

$\left(\frac{6}{2}, \frac{6}{2}\right)$ or (3,3)

Let co-ordinates of D be (x, y), then

$3=\frac{x+6}{2}=6=x+6 \Rightarrow x=6-6=0$

$\Rightarrow y=6-7=-1$

$\therefore$ Co-ordinates of D are (0,-1)

Ans (b)

Question 10

A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, – 5) is the mid-point of PQ, then the coordinates of P and Q are, respectively

(a) (0, – 5) and (2, 0)

(b) (0, 10) and ( – 4, 0)

(c) (0, 4) and ( – 10, 0)

(d) (0, – 10) and (4, 0)

Sol :

A line intersects y-axis at P and x-axis a Q.

R (2, -5) is the mid-point

Let co-ordinates of P be (0, y) and of Q be (x, 0), then

$2=\frac{0+x}{2} \Rightarrow x=4$

and $-5=\frac{y+0}{2} \Rightarrow y=-10$

$\therefore$ Co-ordinates of P are (4,0) and of Q are (0,-10)

Ans (d)

Question 11

The points which divides the line segment joining the points (7, – 6) and (3, 4) in the ratio 1 : 2 internally lies in the

(a) Ist quadrant

(b) IInd quadrant

(c) IIIrd quadrant

(d) IVth quadrant

Sol :

A point divides line segment joining the points

A (7, -6) and B (3, 4) in the ratio 1 : 2 internally.

Let (x, y) divides it in the ratio 1 : 2

$\therefore x=\frac{m x_{2}+n x_{1}}{m+n}$

$=\frac{1 \times 3+2 \times 7}{1+2}=\frac{3+14}{3}$

$=\frac{17}{3}=5 \frac{2}{3}$

$y=\frac{m y_{2}+n y_{1}}{m+n}=\frac{1 \times 4+2 \times(-6)}{1+2}$

$=\frac{4-12}{3}=\frac{-8}{3}$

We see that x is positive and y is negative.

$\therefore$ It lies in the fourth quadrant.

Ans (d)

Question 12

The centroid of the triangle whose vertices are (3, – 7), ( – 8, 6) and (5, 10) is

(a) (0, 9)

(b) (0, 3)

(c) (1, 3)

(d) (3, 3)

Sol :

Centroid of the triangle whose Vertices are (3, -7), (-8, 6) and (5, 10) is

$\left(\frac{3-8+5}{3}, \frac{-7+6+10}{3}\right)$ or $\left(0, \frac{9}{3}\right)$

or (0, 3)

Ans (b)