How do I find the equation of a line connecting points a(p,q) and b(r,s)?

First we need to find the gradient of the line connecting points a and b:
gradient m = (change in y)/(change in x) = (q - s)/(p -r)

Now we use the following equation:

y - y1 = m(x - x1)

substituting suitable values for (x1, y1) (can be points a or b but we'll use point a this time) and m (calculated above):

Using point a:

y - q = [(q-s)/(p-r)](x - p)

and so the equation in the form y = f(x) is:
y = [(q-s)/(p-r)]x + (q-s)/(p-r) + q

Answered by Chris W. Maths tutor

5076 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Show that (2x^2 + x -15)/(2x^3 +6x^2) * 6x^3/(2x^2 - 11x + 15) simplifies to ax/(x + b) where a and b are integers


For all values of x, f(x)=(x+1)^2 and g(x)=2(x-1). Show that gf(x)=2x(x+2)


Solve the simultaneous equations..... 3x - y + 3 = 11 & 2x^2 + y^2 + 3 = 102 where X and Y are both positive integers.


Which has greater area? A parallelogram with base length 10cm and perpendicular height 6cm, or a circle of diameter 8cm.


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