Given two coordinate points (a1,b1) and (a2,b2), how do I find the equation of the straight line between them?

Firstly, remember that we can write such a line in the form y = mx + c, where x and y are our two axes, m is the gradient of the line, and c is the y-axis intersection point. Given this general formula, we need to use our two co-ordinates to work out two values: m and c. Let's start with m. We can think of m as the amount we go up or down (the y-axis) when we take one step forward (along the x-axis). In other words, m is simply the vertical difference between our two coordinates divided by the horizontal difference: m = (b2 - b1)/(a2 - a1). Now what about c? Now we only have one 'unknown' in our equation y = mx + c, and because we know both our points (a1,b1) and (a2,b2) are on the line, that means they satisfy the equation. Therefore, we can plug in either (a1,b1) or (a2,b2) into y = mx + c and solve for c. So: b1 = ma1 + c, and we get c = b1/(ma1). Equally, we could have used c = b2/(m*a2), and it can be a good double check to see if they're the same value. Now we have both m and c, and so our equation is: y = ((b2 - b1)/(a2 - a1))x + b1/(ma1).

Answered by Lewis H. Maths tutor

6284 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

If y=cos(3x)cosec(4x), find dy/dx.


Prove the change of base formula for logarithms. That is, prove that log_a (x) = log_b (x) / log_b (a).


Given that Sin(A) = 1/sqrt(3), show that Tan(A) = 1/sqrt(2)


find dy/dx for the equation y = 6x ^(1/2)+x+3


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