Answer to Question #160106 in Algebra for ciarah simpson

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?

Answer 1

All the streets are either perpendicular or parallel to other Streets so the given streets equation -7x+3y= - 21.5 is also either perpendicular or parallel to the other Streets i.e. also perpendicular and parallel to street PQ. So that for street PQ

1. If PQ is parallel to AB, then

Slopes of both Streets are equal.

-7x+3y = -21.5 compared it to y = mx + c

Where 'm' is slope.

$y = \frac{7x}{3}-\frac{21.5}{3}$

Thus slope of AB is 7/3 and it is also the same slope of PQ. So equation of PQ is

$y = \frac{7x}{3} + c$

Where 'c' is y- axis intercept.

Now for centre Street PQ let the PQ pass through origin (0,0). Thus it satisfy equation of PQ.

0 = 7/3 (0) + c

c = 0, y-intercept is zero

Therefore the street which is parallel to street PQ is

y = 7x/3 or 7x - 3y = 0.

2. But if PQ is perpendicular to AB then product of both Street's slope gives -1.

Slope of PQ × slope of AB = -1

m × (7/3) = -1

m = (-3/7)

Thus equation of PQ perpendicular to AB is

$y = mx + c \\ y = \frac{-3x}{7} + c$

This street PQ pass through center i.e origin (0,0) so this point satisfy the above equation. Thus we get

0 = (-3/7) × 0 + c

c = 0

So PQ is

$y = \frac{-3x}{7}$ or 3x + 7y = 0 .