(i) Find the gradient of the straight line passing through the points: (0,3) and (9,21). (ii) Write down the equation of the line in form y = mx + c

(i) To find the gradient of a straight light, we take any two (different) points on the straight line and compute the change in Y divided by the change in X. So here this is; (21-3)/(9-0) = 18/9 = 2. So the grandient is +2. (ii) To put the straight line into the form y=mx+c, we first note that 'm' is the gradient, and so is 2. Then, we substitute values for 'y' and 'x' using any one of our points. So at the point '(0,3)' we have x=0 and y=3. So we have 3=0*2 + c, so c =3. Therefore we have y=2x +3!

CG
Answered by Charlie G. Maths tutor

5261 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

ABCD is a rhombus on a graph. B(7,10). AC: y=7-4x. Find an equation for DB in the form tx+py+r=0 where t,p&r are integers.


A rectangular path has perimeter of 240m. If the rectangle is split lengthways, two paths of 160m are formed. Work out the lengths of the sides of the original path.


Solve these simultaneous equations; 2y + x = 8, 4 + 4y = 8x


find the coordinates of the single stationary point of the curve with equation y=8x^2 + 2/x


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning