Write the equation of the line that has a slope -1/3 and passes through (-6,5).

Answer 1

We are given m (slope) = -1/3 and the point (-6, 5) on the line.

Let's use y = mx + b which is called the slope-intercept form for the equation of a line where b is the y-intercept (the point at which the line intersects the y-axis.

So in y = mx + b replace x with -6, y with 5 and m with -1/3.

We get 5 = -1/3(-6) + b Now solve for b by first simplifying -1/3(-6) and we get,

5 = 2 + b Now subtract 2 from both sides

-2 -2

3 = b

Finally, in the y = mx + b, replace m with -1/3, b with 3, and the equation of the line we are looking for becomes y = -1/3x + 3

Answer 2

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

m = -1/3 (x1,y1) = (-6,5)

y - 5 = -1/3(x-(-6))

multiplying by 3 both sides to remove fraction

3 (y-5) = -1( x+6)

3y - 15 = -x - 6

Adding x and 15 both sides

x+3y = 9

Answer 3

One way of graphing a line is to use the slope-intercept formula. This linear equation (formula to graph a line) is written as:

y = mx + b

Where m is the slope of the line and b is the y-coordinate of the y-intercept.

We already know that the slope of this line is -1/3, so in order to find the y-intercept we plug the point (-6,5) on the line into the equation.

(5) = (-1/3)(-6) + b

5 = 6/3 + b

5 = 2 + b

3 = b

So the answer to your questions is that the equation for a line with slope -1/3 going through point (-6,5) is:

y = -1/3x + 3