1. Reminder – this slide from previous lesson What is the equation of the line through the point (2, 7) with a gradient of ¾? y – y 1 = m(x – x 1 ) y – 7 = ¾(x – 2) IMPORTANT – Clear the fraction by multiplying both sides by 4 4y – 28 = 3(x – 2) 4y – 28 = 3x – 6 General form (all integer values)

2. The two–point method If we are given two points we can find the equation of the line passing through them using what we already know. We start by calculating m the gradient then directly use the formula above. Finding the equation of a line through two points Point – Gradient formula y – y 1 = m(x – x 1 ) Where m = gradient And (x 1, y 1 ) is a point on the line. EXAMPLE:What is the equation of the line through the points (3, 7) and (8, –3)?

3. Example What is the equation of the line through the points (3, 7) and (8, –3)? Now we use the point-gradient formula y – y 1 = m(x – x 1 ) y – 7 = –2(x – 3) y – 7 = –2x + 6 y = –2x + 13 or in general form 2x + y – 13 = 0

4. Example The line through the points (– 2, 5) and (4, –3) meets the x-axis at point P. Calculate the coordinates of point P. Now we use the point-gradient formula REMEMBER: Never multiply out these brackets Multiply both sides by 3 to clear the fraction

5. Now we use the point-gradient formula REMEMBER: Never multiply out these brackets Multiply both sides by 3 to clear the fraction Equation of line in general form is The line through the points (– 2, 5) and (4, –3) meets the x-axis at point P. Calculate the coordinates of point P. Where line meets x axis, y coordinate is 0 Sub for y = 0 and solve for x Coordinates of P are (, 0 ) PAGE 83 5D

6. Hints Question 7: Find where the lines intersect by solving the two line equations simultaneously. Use the normal method, elimination – (remember we only use substitution for one quadratic / one linear) Question 8: As above Question 9: Three separate point-gradient calculations Question 10: Find the two line equations and solve simultaneously to locate the (x, y) co-ordinates of the point of intersection.