# C1: The Equation of a Straight Line, lesson 2

## Presentation on theme: "C1: The Equation of a Straight Line, lesson 2"— Presentation transcript:

C1: The Equation of a Straight Line, lesson 2
Learning Objective : to be able to find the equation of a straight line in the form y – y1 = m(x - x1)

Starter: Write the equation of the line y = ½ x + 5 in the form ax + by + c = 0. A line, parallel to the line 6x + 3y – 2 = 0, passes through the point (0,3). Work out the equation of the line.

Finding the equation of a line
Suppose a line passes through A(x1, y1) with gradient m. Let P(x, y) be any other point on the line. x y A(x1, y1) P(x, y) y – y1 x – x1 So This can be rearranged to give y – y1 = m(x – x1). In general: This result should be memorized. The equation of a line through A(x1, y1) with gradient m is y – y1 = m(x – x1)

Finding the equation of a line
Finding the equation of a line given two points on the line A line passes through the points A(3, –2) and B(5, 4). What is the equation of the line? x y B(5, 4) A(3, –2) The gradient of AB, m = The equation of this line could also be found by using the given points to find the gradient of the line and using the equation y – y1 = m(x – x1). The gradient could also be substituted into the equation y = mx + c. The value of c can then be found by substituting the values of x and y given by one of the points on the line. Therefore the gradient of AB is 3 .

Finding the equation of a line
Substituting m = 3 and a co-ordinate pair, A(3, –2), into the equation y – y1 = m(x – x1) gives y + 2 = 3(x – 3) y + 2 = 3x – 9 y = 3x – 11 So, the equation of the line passing through the points A(3, –2) and B(5, 4) is y = 3x – 11.

Example Find the equation of the line with gradient ½ that passes through (-1, -4). Using y – y1 = m (x – x1) y- (-4) = ½ ( x – (-1)) y + 4 = ½ x + ½ y = ½ x - 3 ½ 2y – x + 7 = 0

Task 1 : Work out the gradient of the line joining these pairs of points
(4, 2) (6, 3) (-4, 5) (1, 2) (-3, 4) (7, -6) (1/4, ½) (1/2, 2/3) (3b, -2b) (7b, 2b) The line joining (3, -5) to (6, a) has a gradient of 4. Calculate the value of a.

Task 2 : Find the equation of the line when:
m = 2, (x1, y1) = ( 2, 5) m = -1, (x1, y1) = ( 3, -6) m = ½, (x1, y1) = ( -4, 10) (x1, y1) = ( 2, 4), (x2, y2) = ( 3, 8) (x1, y1) = ( 0, 2), (x2, y2) = ( 3, 5) (x1, y1) = ( 5, -3), (x2, y2) = ( 7, 5) (x1, y1) = ( -1, -5), (x2, y2) = ( -3, 3) (x1, y1) = ( 1/3, 2/5), (x2, y2) = ( 2/3, 4/5)

Task 3 The line y = 4x -8 meets the x-axis at the point A. Find the equation of the line with gradient 3 that passes through the point A. The line y = ½ x + 6 meets the x-axis at point B. Find the equation of the line with gradient 2/3 that passes through the point B. Write your answer in the form ax + by + c = 0 where a, b and c are integers. The lines y = x - 5 and y = 3x – 13 intersect at the point C. The point D has co-ordinates (-4, 2). Find the equation of the line that passes through the points C and D.