# How To Solve Coordinate Geometry Questions Quickly

## Ways to solve Coordinate Geometry Problems Quickly

In coordinate geometry , points are placed on the coordinate plane . There are  two scales , one is  running across the plane called the x-axis  and another scale is a right angles to it called the y – axis

### Solve Coordinate Geometry Questions Quickly

• A coordinate geometry is a branch of geometry where the position of the points on the plane is defined with the help of an ordered pair of numbers also known as coordinates.

### Type 1: How To Solve Quickly Coordinate Geometry -Distance and Equation of Straight Line

Question 1: Find the distance between points A(–2,-5) and B(6,1)?

Options:

A. 100

B. 10

C. 20

D. 5

Solution:    AB = $\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$

AB =$\sqrt{(6- (-2))^2 + (1- (-5))^2}$

AB = $\sqrt{(6+2)^2 + (1+5)^2}$

AB = $\sqrt{(8)^2 + (6)^2}$

AB = $\sqrt{64 + 36}$

AB =$\sqrt{100}$

AB = 10

Correct option: B

Question 2: Find the equation of line whose end points are (4, 5) and (2, 8).

Options:

A. 3x – 2y -2 = 0

B. 3x + 2y -2 = 0

C. 3x + 2y + 2 = 0

D. 3x – 2y + 2 = 0

Solution:    We know that,

y – y1 = m(x – x1)

m = $\frac{(y_{1}- y_{2})}{(x_{1}- x_{2}}$

Therefore, y – y1 = $\frac{(y_{1}- y_{2})}{(x_{1}- x_{2})}$ × (x – x1)

y – 5 = $\frac{(8- 5)}{(4 – 2)}$ × x – 4

y – 5 =$\frac{3}{2}$ ×( x – 4)

2y – 10 = 3x – 12

3x – 2y -2 = 0

Correct option: A

Question 3: Two points (a + 3, b + k) and (a, b) are on the line x – 2y + 9 = 0. Find the value of k.

Options:

A. $\frac{1}{2}$

B. $\frac{2}{3}$

C.$\frac{3}{4}$

D.$\frac{3}{2}$

Solution:    x – 2y + 9 = 0

y = $\frac{x}{2}$+ $\frac{9}{2}$

Slope of the line = $\frac{1}{2}$

Slope of the line using two points = b + k – $\frac{b}{a}$ + 3 –a
= $\frac{k}{3}$

$\frac{k}{3}$ = $\frac{1}{2}$

k = $\frac{3}{2}$

Correct option: D

### Type 2: How To Solve Coordinate Geometry Quickly. In which quadrant does the point lie?

Question 1:  In which quadrant does the point (-7, 3) lie?

Options:

Solution:    The point is negative in the x axis and positive for the y axis, thus the point must lie in the 2nd quadrant.

Correct option: B

Question 2: In which quadrant does the point (2, 3) lie?

Options:

Solution:    Both points are positive. Therefore they will lie in 1st quadrant

Correct option: A

Question 3: In which quadrant does the point (-10, -3) lie?

Options:

Solution:    Both points are negative. Therefore they will lie in 3th quadrant.

Correct option: C

### Type 3: Solve Quickly Coordinate Geometry Questions. Find the Coordinates

Question 1: Find the co-ordinates of the centroid of a triangle whose vertices are (0, 4), (6, 10) and (9, 4).

Options:

A. 6, 5

B. 7, 5

C. 5, 6

D. 5, 7

Solution:     We know that, Centroid of a triangle with its vertices (x1,y1), (x2 ,y2), (x3,y3)

C =$\frac{x_{1}+ x_{2} + x_{3}}{3}$, $\frac{y_{1}+ y_{2} + y_{3}}{3}$

C =$\frac{0+ 6 + 9}{3}$, $\frac{4+ 10 + 4}{3}$

C =$\frac{15}{3}$, $\frac{18}{3}$

C = 5, 6

Correct option: C

Question 2: If the distance between two points A(a, –3) and B(3, a) is 6 unit, then a = ?

Options:

A. ± 3

B. – 5

C. 4

D. 0

Solution:    We know that, Distance between two points A(x1, y1) and B(x2, y2)

AB = $\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$

AB = $\sqrt{(3- a)^2 + (a+ 3)^2}$

AB = √2a² + 18

According to question,

$\sqrt{2}$a²+18 = 6

2a² + 18 = 36

a² = 9

a = ±3

Correct option: A

Question 3: The line passing through (4,3) and (y,0) is parallel to the line passing through (–1,–2) and (3,0). Find the value of y?

Options:

A. 2

B. -2

C. -1

D. 1

Solution:    Slope of line passing through (4,3) (y,0)

m = $\frac{(y_{1}- y_{2})}{(x_{1}- x_{2})}$

$m_{1}= \frac{(3- 0)}{(4- y)}$

Slope of line passing through (–1,–2) (3,0)

$m_{2}= \frac{(-2 – 0)}{(-1- 3)}$

If two lines are parallel then, there slopes are equal

$m_{1}= m_{2}$

$\frac{3}{4- y }$ = $\frac{-2}{-4}$

$\frac{3}{4- y }$ = $\frac{2}{4}$

8 – 2y = 12

– 2y = 12 – 8

– 2y = 4

y = $\frac{4}{-2}$

y = – 2

Correct option: B

### Type 4: Coordinate Geometry Solve Questions Quickly.Find the area

Question 1: A(a,b), B(c,d) and C(e,f) are the vertices of a triangle. Given that:

Option 1: AB + BC > AC

Option 2: Area of the triangle = $\frac{1}{2} ×[ (a(c-e) + (c(f-b) ) + (e(b-d) )]$

Which of the following are true?

Options:

A. Option I is true

B. Option II is true

C. Both are true

D. None of the above

Solution:    The formula given in option II is wrong.

The correct formula is

A = $\frac{1}{2} ×[ (a(d-f) + (c(f-b) ) + (e(b-d) )]$

Correct option: A

Question 2: Find the area of the triangle formed by the vertices (31, 25), (23, 30) and (33, 28)

Options:

A. 25

B. 15

C. 19

D. 17

Solution:    Area of triangle = $\frac{1}{2} ×[ (x_{1}(y_{2}-y_{3}) + (x_{2}(y_{3}-y_{1}) ) + (x_{3}(y_{1}-y_{2}) )]$

A=$\frac{1}{2} ×[ (31(30-28) + (23(28-25) ) + (33(25-30) )]$

A = 17

Correct option: D

Question 3: Find the area of the triangle formed by the vertices (-8, 19), (-7, 31) and (13, 25)

Options:

A. 125

B. 135

C. 119

D. 123

Solution:    Area of triangle = $\frac{1}{2} ×[ (x_{1}(y_{2}-y_{3}) + (x_{2}(y_{3}-y_{1}) ) + (x_{3}(y_{1}-y_{2}) )]$

A= $\frac{1}{2} ×[ (- 8 (31-25) + (7(25 – 19) ) + (13(19-31) )]$

A = 123

Correct option: D

Read Also – Formulas for Coordinate geometry

More Practice Questions of various Topics