Write the equation of the line when m = 4 and the point (3, 15) lies on the line.

y=4x+3

Step-by-step explanation:

y=mx+b

15=4(3)+b

15=12+b

-12 -12

3=b

y=4x+3

The desired point is thus (-1/2, -5).

Step-by-step explanation:

The x-component of this directed line segment is 2 - (-8), or 10, and the y-component is -7 - (1), or -8.  This segment is in Quadrant II, since the x-component is positive and the y-component is negative.

The point of interest is (3/4) of the way in the positive x-direction from x = -8.  We can express this symbolically as -8 + (3/4)(10), or -8 + 7.5, or -1/2.

The point of interest is 3/4 of the way in the negative y direction from 1, or:

1 + (3/4)(-8), or 1 - 6, or -5.

The desired point is thus (-1/2, -5).