hayyleeyn1234
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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?

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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 L class=

Sagot :

DarkAngelBaby
its B -1.5x − 3.5y = -31.5
calculista

Answer:

[tex]-1.5x-3.5y=-31.5[/tex]

Step-by-step explanation:

we know that

If two lines are perpendicular

then

the product of their slopes is equal to minus one

so

[tex]m1*m2=-1[/tex]

In this problem line AB and line PQ are perpendicular

Step 1

Find the slope of the line AB

The equation of the line AB is

[tex]-7x+3y=-21.5[/tex]

isolate the variable y

[tex]3y=7x-21.5[/tex] ------> [tex]y=(7/3)x-21.5/3[/tex]

The slope of the line AB is equal to

[tex]m1=7/3[/tex]

Step 2

Find the slope of the line PQ

remember that

[tex]m1*m2=-1[/tex]

we have

[tex]m1=7/3[/tex] ----> slope line AB

so

substitute and solve for m2

[tex](7/3)*m2=-1[/tex]

[tex]m2=-3/7[/tex]

Step 3

Find the equation of the line PQ

The equation of the line into point-slope form is equal to

[tex]y-y1=m(x-x1)[/tex]

we have

[tex]m=-3/7[/tex]

[tex]P(7,6)[/tex]

substitute

[tex]y-6=(-3/7)(x-7)[/tex]

[tex]y=(-3/7)x+3+6[/tex]

[tex]y=(-3/7)x+9[/tex] -----> multiply by [tex]7[/tex] both sides

[tex]7y=-3x+63[/tex]

[tex]7y+3x=63[/tex] -----> divide by [tex]2[/tex] both sides

[tex]1.5x+3.5y=31.5[/tex] -----> multiply by [tex]-1[/tex] both sides

[tex]-1.5x-3.5y=-31.5[/tex]

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