Number 15. find the equation of the line that passes through (3,-1) and is perpendicular to the line that passes through (-5, 2) and (-1, 7)

For lines to be perpendicular their slopes must be negative reciprocals of one another, mathematically:

m1*m2=-1

So we first need to find the slope of the reference line.

m=(y2-y1)/(x2-x1)=(7-2)/(-1--5)=5/4

So the perpendicular line will have a slope of:

5m/4=-1

m=-4/5

So our perpendicular line so far is:

y=-4x/5+b, now we can use point (3,-1) to solve for the y-intercept, "b"

-1=-4(3)/5+b

-1=-12/5+b

-5/5+12/5=b

7/5=b

So the line is:

y=-4x/5+7/5 or more neatly

y=(-4x+7)/5

y=-0.8x+1.4