# How To Solve Coordinate Geometry Quickly

## Ways to solve Coordinate Geometry Problems Quickly

In This page we will be discussing about the ways how we can solve Coordinate Geometry Questions Quickly using basic Formulas, Tips and Tricks.

### 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

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