# Write the point slope form of an equation of the line that passes through the given point and has a given slope. (1;3), m=-2 (-6;3), m=-2/3

### 2 Answers | Add Yours

The equation of the line with slope m passing through a point (x1,y1) is given by:

(y-y1) = m(x-x1)...(1)

Therefore, the equation of the line with slope m = -2 and passing through (1,3) is got by substituting m = -2 and (x1,y1) = ( 1,3) in the equation (1):

y-3 = (-2)(x-1) which is in point slope form.

Similarly the the equation of the line with slope m = -2/3 and passing though (x1 , y1) = (-6, 3) is got similarly by substitution in y - y1 = m(x-x1).

y - 3 = (-2/3)(x-(-6)). Or

y - 3 = (-2/3) (x+6) is in the point slope form.

There are two basic forms of an equation of the line, the point-slope and the standard form.

(the standard is sometimes also called the slope-intercept form).

**Standard:**

y = mx+n (1)

where m - the slope of the line and n - the y-intercept (the value of y when x=0)**Point-Slope:**

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

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

We'll substitute the first pair slope-given point in (2):

**Point slope: y - 3 = -2(x - 1)**

We'll remove the brackets:

y - 3 = -2x + 2

We'll put the equation in the standard form by adding 3 both sides:

y = -2x + 2 + 3

**Standard: y = -2x + 5**

We'll do the same thing with the next given pair: (-6;3), m=-2/3.

**Point-Slope**: y - 3 = (-2/3)(x + 6)

We'll remove the brackets:

y - 3 = -2x/3 - 4

We'll add 3 both sides:

y = -2x/3 - 4 + 3

y = -2x/3 - 1

** Standard**: y = -2x/3 - 1