# Co – ordinate geometry Questions and Answers

## Questions and Answers on Coordinate Geometry

This page deals with different types of Coordinate geometry Questions and Answers which can be asked in placement exams or any other competitive exa ms which has included Coordinate Geometry in its Syllabus.

### Co-ordinate Geometry Rules

• When the x and y axis intersect at a point then it is called origin. Both x and y tend to be 0.
• The values are positive on the right hand side of the x-axis and the values are negative on the left hand side of x axis.
• The value on the upper side of the y axis tend to be positive and the value below tend to be negative.
• By a set of two numbers a point on the plane can be located. First value will be of x axis and second value will be of y-axis which will determine the unified position on the plane.

Example 1:

Find the equation of straight line passing through (2, 3) and perpendicular to the line 3x + 2y + 4 = 0

Options:

a. y=5/3x- 2
b. 3Y=2x+5
c. 3Y=5x-2
d. None of these

Solution:

The given line is 3x + 2y + 4 = 0 or y = -3x / 2 – 2
Any line perpendicular to it will have slope = 2 / 3
Thus equation of line through (2, 3) and slope 2 / 3 is
$(y – 3) =\frac{2}{3(x – 2)}$
3y – 9 = 2x – 4
3y – 2x – 5 = 0.

Correct Option is b.

Example 2:

Find the coordinate of the point which will divide the line joining the point (2,4) and (7,9) internally in the ratio 1:2?

Options:

a. (5/3 , 1/3)
b. (3/8 , 3/11)
c. (8/3 , 11/3)
d. (11/3 , 17/3)

The internal division will use the formula
$\frac{(mx_2 + nx_1)}{(m + n)}$
$y = \frac{(my_2 + ny_1)}{(m + n)}$.
So, the point becomes (11/3, 17/3).

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### CO-ordinate Geometry Questions and Answers

1. Given two points A(-2,0) and B(0, 4), M is a point with coordinates (x, x), x ≥ 0. P divides the joining of A & B in the ratio 2: 1. C & D are the midpoints of BM and AM respectively. Area of the ∆AMB is minimum if the coordinates of M are

(0,0)

(0,0)

70.45%

(1,1)

(1,1)

6.82%

(2,2)

(2,2)

15.91%

(2,2)

(2,2)

6.82%

Area of [llatex]\Delta AMB = \frac{1}{2}\begin{vmatrix}
x & x& 1\\
-2 & 0& 1\\
0 & 4& 1
\end{vmatrix}[/latex]
$\left | \frac{1}{2}(-4x + 2x - 8) \right | = \left | -(x + 4) \right |$
which minimum for x = 0 and thus the co-ordinates of M are (0,0)

2. If the centroid of a triangle formed by (7, x), (y, –6), and (9, 10) is (6, 3), then the values of x and y are respectively.

5,2

5,2

34.78%

2,5

2,5

58.7%

1,0

1,0

2.17%

0,0

0,0

4.35%

Centroid = $\left ( \frac{x_1 + x_2 + x_3}{3} \right ), \left ( \frac{y_1 + y_2 + y_3}{3} \right )$
$\Rightarrow (6,3) = \left ( \frac{7 + y + 9}{3} \right ), \left ( \frac{x - 6 + 10}{3} \right )$
$\Rightarrow 6 = \frac{7 + y + 9}{3} \text{ and }3 = \frac{x - 6 + 10}{3}$

y = 2 and x = 5

3. (0, 0, 0) (a, 0, 0), (0, b, 0) and (0, 0, c) are four distinct points. What are the coordinates of the point which is equidistant from the four points?

((a+b+c)/3, (a+b+c)/3, (a+b+c)/3)

((a+b+c)/3, (a+b+c)/3, (a+b+c)/3)

28.26%

(a, b, c)

(a, b, c)

19.57%

(a/3, b/3, c/3)

(a/3, b/3, c/3)

26.09%

(a/2, b/2, c/2)

(a/2, b/2, c/2)

26.09%

$OP = \sqrt{\left ( \frac{a}{2} - 0 \right )^{2} + \left ( \frac{b}{2} - 0 \right )^{2} + \left ( \frac{c}{2} - 0 \right )^{2} } = \sqrt{\frac{a^2 + b^2 + c^2}{4}}$
$OQ = \sqrt{\left ( \frac{a}{2} - a \right )^{2} + \left ( \frac{b}{2} - 0 \right )^{2} + \left ( \frac{c}{2} - 0 \right )^{2} } = \sqrt{\frac{a^2 + b^2 + c^2}{4}}$
$OR = \sqrt{\left ( \frac{a}{2} - 0 \right )^{2} + \left ( \frac{b}{2} - b \right )^{2} + \left ( \frac{c}{2} - 0 \right )^{2} } = \sqrt{\frac{a^2 + b^2 + c^2}{4}}$
$OS = \sqrt{\left ( \frac{a}{2} - 0 \right )^{2} + \left ( \frac{b}{2} - b \right )^{2} + \left ( \frac{c}{2} - c \right )^{2} } = \sqrt{\frac{a^2 + b^2 + c^2}{4}}$

4. What is the ratio in which the point C(-2/7,-20/7) divides the line joining the points A(–2, –2) and B(2, –4)?

1 : 3

1 : 3

7.14%

3:4

3:4

80.95%

1:2

1:2

7.14%

2:3

2:3

4.76%

$C=\frac{2K-2}{K+1},\frac{-4K+2}{K+1}$
[Latex]\frac{2K-2}{K+1}=\frac{-2}{7}\[/latex]
$C(\frac{-2}{7},\frac{-20}{7})$
$\frac{2(K-1)}{K+1} =\frac{-2}{7}$
7K-7 = -K-1
8K =6=K=[Latex]\frac{3}{4}[/latex]
K:1 = $\frac{3}{4}$:1= 3:4

5. The co-ordinates of incentre of ∆ ABC with vertices A(0, 6), B(8, 12), and C(8, 0) is.

(5, 6)

(5, 6)

60%

(–4, 3)

(–4, 3)

12.5%

(8, 11)

(8, 11)

7.5%

(16/3,0)

(16/3,0)

20%

$a = BC = \sqrt{0^{2} + (12 - 0)^{2}} = 12$
$b = AC = \sqrt{(0 - 8)^2 + (6 - 0)^{2}} = 10$
$c = AB = \sqrt{8^2 + 6^2} = 10$
Incenter is
$\left ( \frac{ax_1 + bx_2 + cx_3}{a + b + c} \right ),\left ( \frac{ay_1 + by_2 + cy_3}{a + b + c} \right )$
i.e
$\left ( \frac{12\times 0 + 10\times 8 + 10\times 8}{12 + 10 + 10} \right ), \left ( \frac{12\times 6 + 10\times 12 + 10\times 0}{12 + 10 + 10} \right )$
$\left ( \frac{160}{32}, \frac{192}{32} \right ) = (5,6)$

6. If t₁ ≠ t₂ and the points A (a, 0), B (at₁², 2at₁) and C (at₂², 2at₂) are collinear, then t t is equal to

1

1

18.92%

-1

-1

72.97%

2

2

5.41%

-2

-2

2.7%

$\Delta = \frac{1}{2}\begin{vmatrix}
a & 0& 1\\
at_1^2& 2at_1& 1\\
at_2^2 & 2at_2& 1
\end{vmatrix}$

$= \frac{1}{2}\times (2a)\times a\times \begin{vmatrix}
1 & 0& 1\\
t_1^2 & t_1& 1\\
t_2^2 & t_2& 1
\end{vmatrix}$

$\therefore \Delta = 0 \Rightarrow (t_1 - t_2)+(t_1^2t_2 - t_2^2t_1)= 0$
$\Rightarrow (t_1 - t_2) + t_1t_2(t_1 - t_2) = 0$
$\Rightarrow (t_1 - t_2)(1 + t_1t_2)= 0$
$\Rightarrow t_1t_2 = -1$

7. The area of quadrilateral ABCD whose vertices in order are A(1, 1), B(7, –3), C(12, 2) and D(7, 21) is

132 sq units

132 sq units

64.86%

66 sq units

66 sq units

13.51%

124 sq units

124 sq units

8.11%

86.5 sq units

86.5 sq units

13.51%

$=\frac{1}{2}\begin{vmatrix}
x_1 - x_3 & y_1 - y_2\\
x_2 - x_4 & y_2 - y_3
\end{vmatrix}$

$=\frac{1}{2}\begin{vmatrix}
1 - 12 & 1 - 2\\
7 - 7 & -3 - 21
\end{vmatrix}$

$=\frac{1}{2}\begin{vmatrix}
-11 & -1\\
0 & -24
\end{vmatrix}$

$\frac{1}{2}(264 - 0) = 132$sq.m

8. The vertices of a triangle ABC are A (2, 3, 1), B (–2, 2, 0), and C(0, 1, –1). What is the magnitude of the line joining mid points of the sides AC and BC?

1/√2 unit

1/√2 unit

10.81%

1 unit

1 unit

8.11%

3/√2 unit

3/√2 unit

70.27%

2 unit

2 unit

10.81%

Mid Point of A and C
$\left ( \frac{2+0}{2}, \frac{3 + 1}{2}, \frac{1 - 1}{2} \right ) = (1, 2, 0)$
Mid Point of B and C
$\left ( \frac{-2+0}{2}, \frac{2 + 1}{2}, \frac{0 - 1}{2} \right ) = \left ( -1, \frac{3}{2}, \frac{-1}{2} \right )$
Magnitude
$=\sqrt{(1 + 1)^2 + \left ( 2 - \frac{3}{2} \right )^2 + \left ( \frac{1}{2} \right )^2}$
$=\sqrt{4 + \left ( \frac{2}{4} \right )+ \left ( \frac{1}{4} \right )}=\frac{3}{\sqrt{2}}$

9. What is the reflection of the point (1.6, 5) in the line y = 1.4?

(1.6, 2.2)

(1.6, 2.2)

17.14%

(1.6, –1.4)

(1.6, –1.4)

11.43%

(1.6, –3.6)

(1.6, –3.6)

14.29%

(1.6, –2.2)

(1.6, –2.2)

57.14%

10. A Rod is placed in two-dimensional Co-ordinate plane and co-ordinate of its ends are A=(6, – 4) and B=(0, 8). A carpenter wants to cut this road in 5:1. What are the co-ordinates of that point where he should cut?

(–1, 6)

(–1, 6)

22.22%

(1, 6)

(1, 6)

66.67%

(–1, –6)

(–1, –6)

4.44%

(1, –6)

(1, –6)

6.67%

$X = \frac{5\times 0 + 1\times 6}{6} = 1$
$Y = Y = \frac{5\times 8 + 1\times -6}{6} = 6$
C = (X,Y) = (1,6)

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