Find the equation of a straight line given two of its points (1,3) and (-2,5). Write your answer in the form y = mx + c.

There are two main steps required to find the equation of a straight line: Finding the gradient of the line AND the slope intercept c. The gradient is the value "m" in the equation "y = mx + b". The formula for finding the gradient given two points (x1,y1) and (x2,y2) is GRADIENT=(y2-y1)/(x2-x1). In our case then the gradient=(5-3)/(-2-1)=(-2/3). (Note, it's important to remember that the x and y values have to be subtracted in the same order). Now our equation is y = (-2/3)x + c , the second main step is substituting in one of the points to find c. Let's choose the point (1, 3) as it does not have negatives so it's harder to make a mistake. We substitute this point into our equation which gives us: 3 = (-2/3) * (1) + c, which is just 3 = (-2/3) + c. Now we have to solve for c, to do this we add (2/3) to both sides to get rid of the (-2/3) on the right hand side. 3+(2/3)=(9/3)+(2/3)=(11/3)=c So our final answer is y=(-2/3)x+(11/3)

Answered by Stanislaw C. Maths tutor

4141 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Show that the recurring decimal 0.0151515..=1/66 (4H)


Solve the simultaneous equations: 3x + y =11, 2x + y = 8


How do I apply the four operations to Fractions?


How do you factorise x^2 -4 = 0?


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