Coordinate Geometry(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

COORDINATE GEOMETRY MCQs

Coordinate Geometry

Total Questions : 68 | Page 3 of 7 pages

Question 21. The perpendicular bisector of a line segment AB passes through the origin. If the coordinates of A are (-2, 0) then the distance of point B from the origin is ___ units.

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:
The perpendicular bisector bisects the line AB at (0, 0). The givenpoint A (-2, 0) lies on the negative side ofthe origin so point B lies on the positive side of the origin with coordinates, (2, 0). Hence, the distance from the y-axis on either side of the origin should be same, i.e., 2 units.

Question 22. What are the coordinates of a point on the Y-axis and at a distance of 5 units in the negative direction of y-axis?

(0, 5)
(0,-5)
(5, 0)
(-5, 0)

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Answer: Option B. -> (0,-5)
:
B
The abscissa of apoint lying on the y-axis is 0 and the point on the negative y-axis has a negative ordinate, therefore, the point is (0, -5).

Question 23. Any point on x- axis is in the form (a, 0).

True
False
(0,0); (0,5); (5,3); (3,0)
(0,0); (0,3); (5,3); (3,5)

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Answer: Option A. -> True
:
A
Any point on x- axis is in the form (a, 0).

Question 24. The point of intersection of axes is __.

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Point of intersection of axes is called Origin.

Question 25. The ordinate of all points on the x-axis is ___.

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Any point on the x-axis is of the form (x, 0). So, any point on the x-axishas an ordinate0.

Question 26. If any point P(x,y) is in the II Quadrant then the value of x is ___ and the value of y is positive.

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In the II Quadrant, the value of x is negative and the value of y is positive.

Question 27. Abscissa of a point is positive in __________ .

1st quadrant
4th quadrant
2nd Quadrant
Both 1st quadrant and 4th quadrant

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Answer: Option D. -> Both 1st quadrant and 4th quadrant
:
D
Distance of a point from y-axis is called abscissa.
In order to have positive values of abscissa, the x-values should be positive.
x - values are positive in1^st quadrant and 4^th quadrant.
So,points in 1^st quadrant and 4^th quadrant have positive abscissa.

Question 28.

The points $(a, a), (- a, - a)$ and $(- \sqrt{3} a, \sqrt{3} a)$ are the vertices of a/an _____ triangle.

scalene
right angled
isosceles right angled
equilateral

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Answer: Option D. -> equilateral
:
D

Let's name the points as $A (a, a), B (- a, - a)$ and $C (- \sqrt{3} a, \sqrt{3} a)$ .

By distance formula, distance between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
$B C = \sqrt{(- \sqrt{3} a + a)^{2} + (\sqrt{3} a + a)^{2}} = \sqrt{a^{2} + (\sqrt{3} a)^{2} - 2 \times a \times \sqrt{3} a + \sqrt{3} a)^{2} + a^{2} + 2 \times a \times \sqrt{3} a} B C = \sqrt{8 a^{2}} = 2 \sqrt{2} a units Similarly A B = \sqrt{(a + a)^{2} + (a + a)^{2}} = \sqrt{8 a^{2}} = 2 \sqrt{2} a units A C = \sqrt{(- \sqrt{3} a - a)^{2} + (\sqrt{3} a - a)^{2}} = \sqrt{8 a^{2}} = 2 \sqrt{2} a units$

$We get, A B = B C = A C = 2 \sqrt{2} a units ∴ △ ABC is an equilateral triangle .$

Question 29.

The point C (–1,2) divides the line segment AB in the ratio 3:4, where the coordinates of A is (2, 5). The coordinates of B are___________.

(–5, –2)
(–1, – 2)
(–4, – 2)
(5, – 4)

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Answer: Option A. -> (–5, –2)
:
A

If a point $P (x, y)$ divides a line segment joining $(x_{1}, y_{1}) and (x_{2}, y_{2})$ in the ratio $m : n$ , then the coordinates of P are given by:
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n}$
Let $C (- 1, 2)$ divide the line joining $A (2, 5)$ and $B (x, y)$ in the ratio 3:4.
Then,
$C (\frac{3 x + 8}{7}, \frac{3 y + 20}{7})$ = C(-1, 2)
$\Rightarrow \frac{3 x + 8}{7} = - 1 3 x + 8 = - 7 \Rightarrow x = - 5 \Rightarrow \frac{3 y + 20}{7} = 2 3 y + 20 = 14 \Rightarrow y = - 2$
Thus, the coordinates of B are $B (- 5, - 2) .$

Question 30.

Midpoint of the line joining the points ( $x_{1}, y_{1}$ ) and ( $x_{2}, y_{2}$ ) is $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

True
False
(–4, – 2)
(5, – 4)

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Answer: Option A. -> True
:
A

True. Midpoint of the line joining the points ( $x_{1}, y_{1}$ ) and ( $x_{2}, y_{2}$ ) is $(\frac{1 (x_{1}) + 1 (x_{2})}{1 + 1}, \frac{1 (y_{1}) + 1 (y_{2})}{1 + 1})$ = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .