P has coordinates (3,4), Q has coordinate (a,b), a line perpendicular to PQ has equation 3x+2y=7. Find an expression for b in terms of a

Rearrange the equation of the line perpendicular to PQ to give y = -(3/2)x + 7/2. Gradient of this line = -3/2Using the knowledge that the gradients of a line perpendicular to another line is related by a factor of -1/x we know the gradient of PQ is 2/3The gradient is the (difference in y)/(difference in x) which is (b-4)/(a-3) = 2/3multiply the bottom of the LHS to get b-4 = 2a/3 - 2 and take the 4 over to get b = 2a/3 + 2.

JL
Answered by Jack L. Maths tutor

4367 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

AQA GCSE higher specimen paper 1: Question 28


expand out the bracket (2m - 3)(m + 1).


Expand (2x + 5)(9x - 2).


How many solutions does a quadratic equation have?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences