Find the equation of a straight line that passes through the coordinates (12,-10) and (5,4). Leaving your answer in the form y = mx + c

Finding the gradient (m): The gradient is the change in y-axis over the change in x-axis Δy = -10-4= -14        Δx = 12-5=7 Δy/Δx = -14/7 = -2 Accumilating the equation: The equation of a straight line can be deduced by a simple formula y- ya = m(x- xa)      where a is a coordinate which lies on the lie.

The equation: y - 4 = -2(x - 5) y - 4 = -2x + 10

Therefore the equation of the line: y= -2x+14

MM

Related Maths A Level answers

All answers ▸

The curve C has the equation (x^2)+4xy-8(y^2)+27=0. Find dy/dx in terms of x and y.


Find the equation of the tangent to the curve y = (2x -3)^3 at the point (1, - 1), giving your answer in the form y = mx + c.


How do I choose which term do I differentiate/integrate when I am integrating by parts


ln(2x^2 + 9x – 5) = 1 + ln(x^2 + 2x – 15). Express x in terms of e