Point A has coordinates (-1,3). Point B has coordinates (2,-3). Find the equation of the line L that goes through these two points. Point C has coordinates (0,1). Hence or otherwise, find the equation of the line perpendicular to L that goes through C.

First we use the equation gradient = (y1-y2)/(x1-x2) to find the gradient of the line L, using point A (-1,3) and point B (2,-3). This gives us gradient = (3-(-3)/(-1-2) = (3+3)/(-3) = -2. Now we substitute point A into the equation y = mx +c where m is the gradient we just calculated. So 3 = -2*-1+c, so c = 1, and the line equation is y = -2x+1 To find a perpendicular line we use the 'negative reciprocal' of the gradient which means -1/m to find the gradient of the second line. Now we have another gradient, and another point C so we can find the equation of the new line. The new gradient = -1/(-2) = 1/2. So y = 1/2x + c. Using point C (0,1) : 1 = 1/2*0 + c, so c = 1. The second line has equation y = 1/2x + 1

Answered by Asa G. Maths tutor

2510 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve the simultaneous equations; 3x + 2y = 11 x -4y = 13


Write x^2 – 10x + 12 in the form (x – a)^2 + b , where a and b are integers.


Solve x^2+4x-5=0 by completing the square.


f(x)=2x+c, g(x)=cx+5, fg(x)=6x+d; c and d are constants. Work out the value of d


We're here to help

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

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences