Question

1

What is the point-slope form of a line that has a slope of -4 and passes through point (-3, 1)?
=
oy-(-3) --4(x-1)
y-1--4[*-(-3))
-1-y, = -44-3-X)
3-y,--4(1-x) ​

Answer 1

`-4(x+3) = y - 1`

Explanation:

Given

Slope of line = -4

Point = (-3,1)

Required

Point slope equation of line

To get the equation of line, we make use of the following slope formula

`m = (y - y_1)/(x - x_1)`

Where m = -4 and
`(x_1,y_1) = (-3,1)`

So;
`m = (y - y_1)/(x - x_1)` becomes

`-4 = (y - 1)/(x - (-3))`

`-4 = (y - 1)/(x +3)`

Multiply both sides by x + 3

`-4(x+3) = (y - 1)/(x +3) * (x+3)`

`-4(x+3) = y - 1`

Hence, the point slope form of the line is
`-4(x+3) = y - 1`