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

Solution.

The equation of the line passing through A and B is: $-7x + 3y = -21.5.$

It can be written as: $y=\frac{7}{3}x​−\frac{21.5}{3}. ​$

By comparing with the standard form $y=mx+c,$ the slope of this line is $m=\frac{7}{3}.$

Central street PQ is perpendicular to the line passing through A and B. Product of slopes of perpendicular lines is $m1*m2=-1$.

Slope of the perpendicular line is:$Slope\; of \;perpendicular\; line=\newline = \frac{-1}{slope\;of\;parallel\;line}=\frac{-1}{\frac{7}{3}}=\frac{-3}{\;7}.$

Thus the equation of line is: $y=mx+c; \newline y=\frac{-3}{7}x+c; \newline 7y= -3x+7c; \newline 7y+3x=7c.$

Dividing by 2:

$3.5y+1.5x=3.5c.$

To find c with just this information is impossible.It is needed to check the options of the line with the slope -3/7

on the figure.

The line PQ passes through the point (7,6) on the figure.Hence the equation of the line can be found out, using point slope form of a line: $y−y_1=m(x−x_1).$

Slope is $\frac{-3}{7}$ .

$(x_1,y_1) \;is\; (7,6).$

$y-6=\frac{-3}{7}(x-7); \newline 7(y-6)=-3(x-7); \newline 7y-42=-3x+21; \newline 7y+3x=21+42; \newline 7y+3x=63.$

This is the equation of the central line.

Dividing by 2:

$3.5y+1.5x=31.5.$

Or

$-3.5y-1.5x=-31.5$ is also the equation.

Answer:

$3.5y+1.5x=31.5$

or

$-3.5y-1.5x=-31.5$.