# Formulas for Coordinate Geometry

## Coordinate Geometry Formulas

Co-ordinate geometry is considered as one of the most easiest chapters from Quantitative Aptitude Section which can be asked in an Exam. This page here will give you all the required Formulas for Coordinate Geometry.

So that you can ace the questions asked from this chapter in any of the exams you wish to appear.

### What is Coordinate Geometry?

• 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.

### Coordinate Geometry Formulas and Basic Concept

• The point of intersection of the x and the y-axis is known as the origin. At this point, both x and y are 0.
• The values on the right-hand side of the x-axis are positive and the values on the left-hand side of the x-axis are negative.
• Similarly, on the y-axis, the values located above the origin are positive and the values located below the origin are negative.

### Formulas Required  for Solving Coordinate Geometry Questions.

• Distance between two points A(x1, y1) and B(x2, y2)
AB = $\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$
• Slope of line when two points are given (x1, y1) and (x2, y2)
m = $\frac{(y_{1}- y_{2})}{(x_{1}- x_{2})}$
• Slope of line when linear equation is given ax + by = c => $– \frac{a}{b}$
• Midpoint =$\frac{x_{1} + x_{2}}{2},\frac{y_{1} + y_{2}}{2}$
• The co-ordinates of a point R(x,y) that divides a line segment joining two points A(x1, y1) and B(x2, y2) internally in the ratio m:n is given by
x = $\frac{ m x_{2} + nx_{1}}{m + n}$

y = $\frac{my_{2} + ny_{1} }{m + n}$

• The co-ordinates of a point R(x,y) that divides a line segment joining two points A(x1, y1) and B(x2, y2) externally in the ratio m:n is given by
x =$\frac{ m x_{2} – nx_{1}}{m – n}$

y = $\frac{my_{2} – ny_{1} }{m – n}$

• Centroid of a triangle with its vertices (x1,y1), (x2,y2), (x3,y3)
C = $\frac{x_{1}+ x_{2} + x_{3}}{3}$, $\frac{y_{1}+ y_{2} + y_{3}}{3}$
• Area of a Triangle with its vertices A(x1,y1), B(x2,y2), C(x3,y3)
A =$\frac{1}{2} ×[ (x_{1}(y_{2}-y_{3}) + (x_{2}(y_{3}-y_{1}) ) + (x_{3}(y_{1}-y_{2}) )]$
• Division of a line segment by a point
If a point p(x, y) divides the join of A(x1, y1) and B(x2, y2), in the ratio m: n, then
x= $\frac{ m x_{2} + nx_{1}}{m + n}$ and y= $\frac{my_{2} + ny_{1} }{m + n}$
• The equation of a line in slope intercept form is Y= mx+ c, where m is its slope.
The equation of a line which has gradient m and which passes through the point (x1, y1) is =
y – y1 = m(x – x1).

## Some Examples Using Above Formulas:

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Question 1: Given two points A(2, 5) and B(6, 9), find the Distance between points A(2, 5) and B(6, 9)

Answer: Distance between points A(2, 5) and B(6, 9):
Distance = $\sqrt{6-2}^{2}$+$\sqrt{9-5}^{2}$
=$\sqrt{4^{2}+4^{2}}$ =
$\sqrt{32}$ =
5.66 Ans

Question 2:If the slope of a line is -2/3 and it passes through the point (2, 5), find the equation of the line in point-slope form.
Answer: Equation of the line passing through points P(3, 8) and Q(5, -2):
Slope (m) =$\frac{change in y}{change in x}$ =$\frac{-2-8}{5-3}$=$\frac{-10}{2}$=-5
Using the point-slope form: y – y1 = m(x – x1)
y – 8 = -5(x – 3)
y – 8 = -5x + 15
y = -5x + 23

Question 3: A rectangle has vertices at points A(1, 1), B(5, 1), C(5, 3), and D(1, 3). Determine its area and perimeter.

Answer:Rectangle with vertices at points A(1, 1), B(5, 1), C(5, 3), and D(1, 3):
Area = Length * Width Length =
Distance between points B and C:
Length = √((5 – 5)^2 + (3 – 1)^2) = √(0 + 4) = √4 = 2
Width = Distance between points A and B:
Width = √((5 – 1)^2 + (1 – 1)^2) = √(16 + 0) = √16 = 4 Area = 2 * 4 = 8 square units
Question 4: Determine the equation of the circle with center C(2, -1) and a radius of 5.

The equation of a circle with center (h, k) and radius r is $(x-h)^{2} + (y-k)^{2} = r^{2}$
For this circle: = $(x-h)^{2} + (y-(-1)^{2}$ = $r^{2}$=$5^2 (x – 2)^2 + (y + 1)^2$ = 25
Question 5: Given the points A(3, 6), B(8, 6), and C(5, 2), find the area of the triangle ABC.

Answer: Area of triangle ABC with vertices A(3, 6), B(8, 6), and C(5, 2):
Area = ½ |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|
Area = ½ |3(6 – 2) + 8(2 – 6) + 5(6 – 6)|
Area = ½ |12 – 24 + 0|
Area = ½ |-12|
Area = 6 square units

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