# What is the point slope form of the equation of the line that passes through the point (0;2) and it has the slope m = -3 ?

*print*Print*list*Cite

The line passes through the point (0,2) and the slope m = -3.

Let us write the equation in the standard form:

y-y1 = m(x-x1) such that:

(x1,y1) is any point that passes through the line and m is the slope.

Given the point (0,2) and slope m= -3 , we will subsitute these values into the equation:

y-y1= m(x-x1)

y- 2 = -3 (x - 0)

y-2 = -3x

Now add 2 to both sides:

**==> y= -3x + 2 ****is the formula for the line.**

The easiest way to write an equation is in point-slope form. The formula for point-slope is: y - y1 = m(x - x1) where (x1,y1) is a point on your line and m is slope.

Plug in our given values: (0;2) and it has the slope m = -3

y - 2 = -3(x - 0)

If you want simply further.

If a line has a slope m and passes through a point (x1, y1), then its equation in point slope form is given by:

y-y1 = m(x-x1)

Since the given line has a slope m = -3 and (x1, y1) = (0,2), substituting these values into y -y1 = m(x-x1) gives us:

y - 2 = (-3)(x-0) ==> the equation **in point slope form**.

There are two basic forms of an equation for a line: the point-slope and the standard form (the standard is sometimes also called the slope-intercept form).

**The Point-Slope form of an equation:**

(y-y1) = m(x-x1) (2)

where m is the slope and (x1,y1) is the given point.

We'll substitute the slope and given point in (2):

** y - 2 = -3(x - 0)**

We'll remove the brackets:

y - 2 = -3x + 0

We'll put the equation in the standard form by adding 2 both sides:

y = -3x + 2

**Standard form of the equation is: y = -3x + 2**