Question

5. Find the value(s) of x so that the line containing the points (2x + 3, x + 2) and (0, 2) is

perpendicular to the line containing the points (x + 2, -3-3x) and (8, -1).

Answer 1

x = -2 or -9

Explanation:

You want the values of x such that the line defined by the two points (2x+3, x+2) and (0, 2) is perpendicular to the line defined by the two points (x+2, -3-3x) and (8, -1).

Slope

The slope of a line is given by the slope formula:

m = (y2 -y1)/(x2 -x1)

Using the formula, the slopes of the two lines are ...

m1 = (2 -(x+2))/(0 -(2x+3)) = (-x)/(-2x-3) = x/(2x +3)

and

m2 = (-1 -(-3-3x))/(8 -(x+2)) = (2+3x)/(6 -x)

Perpendicular lines

The slopes of perpendicular lines have product of -1:

`(x)/(2x+3)\cdot(2+3x)/(6-x)=-1\\\\x(3x+2)=(2x+3)(x-6)\qquad\text{multiply by $(2x+3)(6-x)$}\\\\3x^2+2x=2x^2-9x-18\qquad\text{eliminate parentheses}\\\\x^2+11x+18=0\qquad\text{put in standard form}\\\\(x+2)(x+9)=0\qquad\text{factor}`

Solutions

The values of x that satisfy this equation are x = -2 and x = -9. The attached graphs show the lines for each of these cases.