The point P has coordinates (4, 5). The point Q has coordinates (a, b). A line perpendicular to PQ is given by the equation 5x+3y=11. Find an expression for b in terms of a.

First we should find the gradient of the line perpendicular to PQ by rearranging the equation.5x+3y=113y=11-5xy=11/5 -5/3xy= -5/3x + 11gradient is -5/3.We know that the gradients of perpendicular lines multiply together to equal -1. Or, if you prefer, to find the perpendicular gradient, you can 'flip' the first gradient and multiply it by -1. So a gradient of 7/8 is 'flipped' to 8/7 and multiplied by -1 to become -8/7.This means the gradient of PQ is:-5/3 flipped = -3/5, then multiplied by -1 = 3/5.We know that gradient is change in y/change in x. So we need to make an equation we can solve using our new gradient, using the points P and Q:change in y/change in x = 3/55-b/4-a = 3/55(5-b) = 3(4-a)25-5b = 12-3a25+3a = 12+5b13+3a=5bb=3/5 a +13/5

Answered by Staś B. Maths tutor

3289 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How to solve a simultaneous equation?


Solve the simultaneous equations, 3x + y = 10 and x + y = 4.


Bhavin, Max and Imran share 6000 rupees in the ratios 2 : 3 : 7. Imran then gives 3/5 of his share of the money to Bhavin. What percentage of the 6000 rupees does Bhavin now have? Give your answer correct to the nearest whole number.


What is the simplified expression of: 3a - a x 4a + 2a? And what rule do we use to carry out the simplification?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences