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CBSE 10th Maths -  Coordinate Geometry- Multiple choice questions (MCQs)

58376.What will be the reflection of the point (4, 5) about the x-axis in the fourth quadrant?
(4, 5)
(4, -5)
(-4, -5)
(-4, 5)
Explanation:

X-axis will act as a plane mirror, and this point will form an image, following the sign convention, at (4, -5) in the fourth quadrant.

58377.Point P lies on the line 3x + 4y – 12 = 0. If the x-coordinate of P is a, then its y-coordinate is ______.
$\dfrac{(12-3a) }{ 4}$
$\dfrac{(12-4a)}{ 3}$
$\dfrac{(12+3a)}{4}$
$\dfrac{(3a-12)}{4}$
Explanation:

The points on the line should satisfy the equation of the line.

So, the point P (a, y) satisfies the equation.

Hence,

3x+ 4y – 12 = 0

3(a) + 4(y) -12 = 0

4y = 12-3a

Y = $\dfrac{(12-3a) }{ 4}$

58378.If point P lies on the line y = -1, find the following a) Its y-coordinate, b) Its x-coordinate
y = -1, x can be any real number
y = 1, x = -1
x = -2, y = -1
x = -1, y = -1
Explanation:

Since it is given that the point P lies on the line y = -1, its y-coordinate will be -1, and the x-coordinate can be any real number.

58379.The distance of the point (–2, –2) from the origin is
$\sqrt{2}$
8
2$\sqrt{2}$
$\sqrt{9}$
Explanation:

Let the origin be O and the point A be (-2, -2)

Using the distance formula,

$OA^{2}$ = $(2^{2} + 2^{2})$

$OA^{2}$ = 8

OA=$\sqrt{8}$=2$\sqrt{2}$

58380.The distance between the points (a, b) and (–a, –b) is:
2$\sqrt{a^{2}+b^{2}}$
0
$\sqrt{a^{2}+b^{2}}$
$a^{2}+b^{2}$
Explanation:

Let A (a, b) and B (-a,-b) be the two points and ’d’ be the distance between them.

By using the distance formula, we get

d=$\sqrt{(-a(-a))^{2}+(-b(-b))^{2}}$

d=$\sqrt{(2a)^{2}+(2b)^{2}}$

d=2$\sqrt{a^{2}+b^{2}}$

58381.Mid-point of the line segment joining the points (–5, 4) and (9, –8) is:
(-2,2)
(7,-6)
(2,-2)
(-7,6)
Explanation:

Midpoint of a line segment joining (x1, y1) and (x2, y2) is [(x1+ x2)/2, (y1+y2)/2]

∴ Mid-point of the line-segment joining the points (–5, 4) and (9, –8) = [(9-5)/2, (4-8)/2] = (2, -2)

58382.Points A (1, 2) and B (3, 4) are two ends of a line segment. Find the point which divides AB in the ratio 3:4.
(4,3)
(2,3)
15/7, 22/7
13/7, 20/7
Explanation:

The coordinates for the point that divides a line in the ratio m:n is

$\dfrac{n\times x_{1}+m\times x_{2}}{m+n}$ , $\dfrac{n\times y_{1}+m\times y_{2}}{n+m}$

Substituting in the equation, we get

=$\dfrac{3\times 3+4\times 1}{3+4}$ , $\dfrac{3\times 4+4\times 2}{3+4}$

= 13/7, 20/7

Therefore, the point which divides the line segment in the ratio 3:4 is 13/7, 20/7.

58383.If (1, 2), (4, y), (x, 6), and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
(2, 3)
(4, 3)
(6, 3)
(2, 5)
Explanation:

The given points are: A(1,2), B(4, y), C(x, 6), and D(3,5)

Since the diagonals of a parallelogram bisect each other,

The coordinates of P are:

X= $\dfrac{(x+1)}{ 2}$ = $\dfrac{(3+4)}{2}$

⇒x+1=7⇒x=6

Y = (5+Y)/2= (6+2)/2

⇒5+y=8⇒y=3

∴ The required values of x and y are:

x=6, y=3.

58384.If the points (a, 0), (0, b) and (1, 1) are collinear, then which of the following is true?
(1/a) + (1/b)= 2
(1/a) + (1/b)= 1
(1/a) + (1/b)= 0
(1/a) + (1/b)= 4
Explanation:

Area of a triangle formed by (x1, y1), (x2, y2), (x3, y3) = 1/2 [x1(y2 – y3) + x2(y3 – y1)+ x3(y1 – y2)]

For 3 points to be collinear, the area of the triangle should be zero:

⇒1/2[a (b−1) + 0(1−0) + 1(0−b)] =0

⇒1/2[a (b−1) +1(0−b)] =0

⇒ab=a+b

⇒ (1/a) + (1/b) = 1

58385.If 2 triangles have the same height, the ratio of their areas is equal to
The ratio of any 2 sides
The ratio of their corresponding bases
The ratio of their heights
1
Explanation:

$\dfrac{Area\:\: of\:\: triangle 1 }{Area \:\: of\:\: triangle 2}$

=$\dfrac{ (\dfrac{1}{2} × Base 1× height) }{ (\dfrac{1}{2} × Base 2× height) }$

= $\dfrac{ base 1}{base 2}$

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