Question

A software designer is mapping the streets for a new racing game. All of the streets are depicted as either perpendicular or parallel lines. The equation of the lane passing through A and B is -7x + 3y = -21.5. What is the equation of the central street PQ?
A. -3x + 4y = 3




B. -1.5x − 3.5y = -31.5




C. 2x + y = 20




D. -2.25x + y = -9.75

Answer 1

Answer: Option B. -1.5x-3.5y=-31.5

Solution:

Equation of the lane passing through A and B: -7x+3y=-21.5

If we isolate y, and express the equation in the slope-intercept form y=mx+b, the coefficient of the variable "x" is the slope of this line:

Adding 7x both sides of the equation:

-7x+3y+7x=-21.5+7x

3y=7x-21.5

Dividing both sides of the equation by 3:

(3/3)y=(7/3)x-21.5/3

y=(7/3)x-21.5/3

The slope of this line is m1=7/3 (coefficient of variable x)

This line is perpendicular to the central line PQ, then the product of their slopes must be equal to -1.

Slope of AB: m1=7/3

Slope of PQ: m2

m1 m2 = -1

Replacing m1=7/3

(7/3) m2 = -1

Solving for m2: Multiplying both sides of the equation by 3/4:

(3/7)(7/3) m2 = (3/7)(-1)

m2 = - 3/7

According with the graph we know that the line PQ passes through the point P=(7,6)=(xp, yp)→xp=7, yp=6

Then, using the point-slope equation:

y-y1=m(x-x1); y1=yp=6, m=m2=-3/7, x1=xp=7

Replacing the values:

y-6=(-3/7)(x-7)

Multiplying both sides of the equation by 7:

7(y-6)=7(-3/7)(x-7)

7y-42=(-3)(x-7)

7y-42=-3x+21

Subtracting 7y and 21 both sides of the equation:

7y-42-7y-21=-3x+21-7y-21

-63=-3x-7y

-3x-7y=-63

Dividing all the terms by 2:

-(3/2)x-(7/2)y=-63/2

-1.5x-3.5y=-31.5