Coordinate Geometry
Content
- Distance Between Two Points
- Mid Point
- Gradient of Straight Line
- Collinear Points
- Parallel and Perpendicular Lines
- Equation of Vertical and horizontal Lines
- Equation of a straight line
- What You need To Find Equation of a Straight Line?
Distance Between Two Points
The distance between two points (x1,y1) and (x2,y2) is
Coordinate of Midpoint
Find the coordinate of the midpoint between (2 , 3) and (4 , 7).
Gradient of a Straight Line
Find the gradient of a straight line passing through (3, 4) and (5, 9).
Gradient = (y2 – y1) / (x2 – x1)
= (9 – 4) / (5 – 3)
= 5 / 2
= 2.5
Collinear Points
Three Points A,B,C are Collinear, if they lie on the same straight line.
Gradient AB = Gradient BC.
Find y, if A(2 , 3), B(4 , 5) and C(7 , y) are collinear.
If A, B and C are collinear, Gradient AB = Gradient BC.
(5 – 3) / (4 – 2) = (y – 5) /(7 – 4)
2/2 = (y – 5) /3
1 = (y – 5) /3
1 x 3 = (y – 5)
3 = y – 5
3 + 5 = y
y = 8
Parallel and Perpendicular Lines
If two lines are parallel, then they have equal gradients.
If two lines are perpendicular, then multiplication of their gradients equals – 1.
Equation of Vertical and horizontal Lines
Equation of vertical line is x = a, a is x-intercept.
Equation of horizontal line is y = b, b is y-intercept.
Equation of a straight line
y= mx + c
m is the gradient
c is the y-intercept, where the line cuts the y axis.
What You need To Find Equation of a Straight Line?
- gradient and y-intercept (or)
- Gradient and coordinate of one point on the straight line (or)
- Coordinates of two points on the straight line (or)
- Equation of a Parallel/perpendicular line and
- y-intercept (or)
- Equation of a Parallel/perpendicular line and coordinate of one point on the straight line.
Find Equation of Straight Line with gradient is 2 and passing through (3,2).
y = mx + c
y = 2x + c
Since the line passing through (3,2), substitute x = 3 , y = 2
2 = 2(3) + c
c = – 4
Hence the equation of the line is y = 2x – 4.
Find Equation of Straight Line Passing through (2 , 3) and (4, 6).
Gradient = m = (6-3) / (4-2) = 1.5
y = 1.5x + c
Since the line passing through (2 , 3), substitute x = 2 , y = 3
3 = 1.5(2) + c
c = 0
Hence the equation of the line is y = 1.5x.
Find Equation of Straight Line Parallel to y = 5x + 10 and passing through (2 , 3)
Since the lines are parallel, their gradients are equal.
m = 5
y = 5x + c
Since the line passing through (2 , 3), substitute x = 2, y = 3
3 = 5(2) + c
3 = 10 + c
c = – 7
Hence the equation of the line is y = 5x – 7.
Finding Equation of Straight Line Perpendicular to y = – 3x + 7 and passing through (3 , 2)
Since the lines are perpendicular, m = 1/3
y = (1/3) x+ c
Since the line passing through (3,2), substitute x = 3 , y = 2
2 = (1/3) (3) + c
c = 1
Hence the equation of the line is y = (1/3) x + 1
Multiply by 3, 3y = x + 3.
Question
- Find the gradient of the line 8y = 4x + 5
- Find the value of m if the point (m, m+2) lies on the line 2y = 3x – 9
- A(1,3) and B(10,12). Find (a) Length of AB (b) Mid-point of AB (c) Equation of AB.
- Find the equation of the line parallel to x-axis and passes through the point (2 , 4).
- Find the equation of the line parallel to y-axis and passes through the point (–6 , –3).
- Find the equation of the line that is parallel to y = 2x + 5 and passes through the point (1 , 5).
- Find the equation of the line that is parallel to y = –6x + 12 and passes through the point (6 , 2).