# Determine the equation for a line that passes through the point (-4,2) and has a slope of 3/2.

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The Point-Slope equation has the form:

*y* – *y*1 = *m*(*x* – *x*1) , where m is the slope and x1 and y1 are point coordinates.

You're given a point (-4,2) which corresponds to the x1 and y1 coordinates, so plugging in our values gives the equation:

y - 2 = 3/2 (x-(-4)) which is your equation, but now to simplify:

y - 2 = 3/2(x+4)

y = 3/2(x+4) + 2

y = 3/2x + 12/2 + 2

y = 3/2x + 6 + 2

y = 3/2x + 8

So for any value of x, positive or negative, you can get the value of y, positive or negative. These two values together (x,y) will be a point somewhere on the line.

The equation of a line passing through a point `(x_1, y_1)` and with slope m is:

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

For a line passing through (-4,2) and with slope 3/2, `x_1 = -4` , `y_1 = 2` and m = 3/2

Substituting these values in the formula given earlier.

`(y - 2)/(x - (-4)) = 3/2`

`(y - 2)/(x + 4) = 3/2`

2*(y - 2) = 3*(x + 4)

2y - 4 = 3x + 12

3x - 2y + 16 = 0

The required equation of the line is 3x - 2y + 16 = 0

slope form is mx+b

y = 3/2 (x-(-4)) +2

y= 3/2 (x+4) +2

distribute the 3/2

y = 3/2x + 6 + 2

y = 3/2x + 8

point (-4,2) and has a slope of 3/2.

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.

Plugging in:

y - 2 = 1.5(x - 2)

You can leave it at that or solve for another form.

y= mx + c

since gradient (m)= 3/2

Sub. m=3/2 into equation

y= 3/2x+c

Next, you have both x (-4) and y (2) coordinate, so sub them both into the equation

2= 3/2(-4)+c

2= -6+c

c= 8

**Thus, y= 3/2x+8**