Starter A car club has an annual hill climb competition. A cross section of part of the hill they race on is drawn below. The beginning and end points of the section, in metres from the start of the hill are given as co-ordinate pairs.

Find: The gradient over this section of the hill climb What is the distance of this section of the hill climb in metres to the nearest metre? What are the co-ordinates of the half-way point of this section of the hill climb? m = 1/5 153m (215, 85)

Note 6: Revision of Equations of Lines The gradient/intercept form for the equation of a line is: y = mx + c gradient y-intercept

Plot y = -2/3x + 2 using the gradient and intercept method Example: Plot y = -2/3x + 2 using the gradient and intercept method y-intercept (c) = 2 Gradient (m) = -2 3 Fall is 2 Run is 3 Negative sign means it leans to LEFT Plot the y-intercept Plot the next point by making a triangle, whose fall is 2 and run is 3 Plot another point in the same manner Connect the points with a straight line

Don’t forget arrows & label  Another option: up (+2) then left (-3)  One option: down (-2) then right (+3) Now join the dots  y = -2/3x + 2 Don’t forget arrows & label

Exercises: Plot the following graphs; y = 2x – 3 Y = -⅖x + 2 y = x

The general form of the equation of a line is: ax + by + c = 0 Equations can be rearranged from gradient-intercept form to the general form by performing operations on both sides of the equation. (a is usually positive)

Example: Write the following equation in the general form y = -2/3x + 2 3y = -2x + 6 2x + 3y – 6 = 0 Multiply equation by 3 Move everything to the LHS

Write the following equations in the general form: Exercises: Write the following equations in the general form: y = 4x – 3 Y = -⅖x + 2 y = ⅙x – ⅚ 4x – y - 3 = 0 2x + 5y - 10 = 0 x - 6y - 5 = 0

y – y1 = m (x – x1) Note 7: Finding Equations of Lines If you know the gradient, m and any point, (x1 , y1) then the equation of the line can be worked out using the formula y – y1 = m (x – x1)

Example 1 Find the equation of the line passing through (– 3, 5) and with gradient 4 Step 1 Write m, x1 and y1 m = 4 x1 = –3 y1 = 5 Step 2 Put these values into the formula y – y1 = m(x – x1) y – 5 = 4(x – – 3) Step 3 Remove brackets. Write in general form y – 5 = 4(x + 3) y – 5 = 4x + 12 y = 4x + 17 0 = 4x - y + 17

Find the equation of the line passing through (– 2, -13) and (3, 2) Example 2 (x1, y1) (x2, y2) Step 1 Find m m = = 3 Step 2 Put these values into the formula y – y1 = m(x – x1) y – -13 = 3(x – – 2) Step 3 Remove brackets. Write in general form Y + 13 = 3(x + 2) Y + 13 = 3x + 6 y = 3x - 7 0 = 3x - y - 7

Page 272 Exercise 8F.1