Maths question year 10

Find the equation of a straight line that passes through point (6,2) and has a gradient of 4.

Using the equation : (y - y') = m(x - x')
Here y' = 2 ; x' = 6 ; m = 4

(y - 2) = 4(x - 6)
y - 2 = 4x - 24
4x -y -22 = 0 --------------> Required equation

We can find the equation by using

y - y1 = m (x - x1)

here m = 4 and (x1, y1) = (6,2)

so y - 2 = 4 ( x - 6)
y - 2 = 4x -24
y = 4x -24 +2
y = 4x -22
or 4x - y - 22 = 0

Use point slope form!
y-2= 4(x-6)
Simplify!

y=mx+c
sub in points and gradient 2=(4)(6)+c
c=-22
y=4x-22