# How To Solve Coordinate Geometry Questions Quickly

## 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.

Read Also –  Formulas to solve co-ordinate geometry

### How To Solve Coordinate Geometry Questions Quickly:  ## 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 = √(x2-x1 )2+(y2-y1 )2
AB = √(6-(-2))2+(1-(-5))2
AB = √(6+2))2+(1+5)2
AB = √(8)2+(6)2
AB = √64+36
AB = √100
AB = 10

### 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 = (y1-y2)/(x1-x2 )
Therefore, y – y1 = (y1-y2)/(x1– x2) *(x – x1)
y – 5 = (8-5)/(4-2 ) * x – 4
y – 5 = 3/(2 ) * x – 4
2y – 10 = 3x – 12
3x – 2y -2 = 0

### 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. 1/2
B. 2/3
C. 3/4
D. 3/2

#### Solution:

x – 2y + 9 = 0
y = x/2 + 9/2
Slope of the line = 1/2
Slope of the line using two points = b + k – b/a + 3 –a
= k/3
k/3 = 1/2
k = 3/2

## 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.

### Question 2:

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

#### Solution:

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

### Question 3:

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

#### Solution:

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

## 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 =(x1+x2+x3)/3, (y1+y2+y3)/3
C =(0+6+9)/3, (4+10+4)/3
C =15/3, 18/3
C = 5, 6

### 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 = √((x2-x1 )2 )+(y2-y1 )2
AB = √(3 – a)² + (a + 3)²
AB = √2a² + 18
According to question, √2a²+18 = 6
2a² + 18 = 36
a² = 9
a = ±3

### 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 = (y1-y2)/(x1-x2 )
m1= 3 – 0/4 – y

Slope of line passing through (–1,–2) (3,0)
m2 = (y1-y2)/(x1-x2 )

m2 = -2 – 0/-1-3

If two lines are parallel then, there slopes are equal
m1 = m2

3/(4-y) = (-2)/(- 4)
3/(4-y) = 2/( 4)
8 – 2y = 12
– 2y = 12 – 8
– 2y = 4
y = 4/-2
y = – 2

## 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 = (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 = ½ ((a(d – f) + c(f – b) + e(b – d))

### 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 = 1/2 ((x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2))
A= 1/2 ((31(30 – 28) + 23(28-25) + 33(25 – 30))
A = 17

### 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 = 1/2 ((x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2))
A= 1/2 ((-8(31 – 25) + -7(25-19) + 13(19 – 31))
A = 123

#### Correct option: D

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