Find the equation of the straight line which passes through the points (3,1) and (-1,3)?

The equation of a line is y=mx+c. The first step is to find m, which is the gradient. This is found by dividing the change in the y values, by the change in x values. So in this case, the gradient would be (1-3)/(3-(-1)) = -2/4 = -1/2. Therefore we can substitue the m in the equation with -1/2: y=(-1/2)x +c. To find c, we substitute the values of one of the points given in the question, into the equation of the line. For example we could choose (3,1) and swap the x in the equation for 3, and the y for 1: 1 = (-1/2)(3) + c. The value of c can now be found by making it the subject of the equation: c=1-(-3/2) = 5/2. Therefore, the equation of the line is: y = (-1/2)x + 5/2

JP
Answered by Joel P. Maths tutor

10093 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

What is the area of a circle with a diameter of 3cm? (To 2 d.p.)


Find the equation of a line which goes through the points (1,0) and (2,5) in the form of y=mx+c


Multiply out and simplify (x+7)(x-3)


s^2 - 2s - 24 = 0


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences