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line p contains point (6 ,- 5 and is perpendicular to line q the equation for line q is y=3x+5 write an equation for line q the find slope of line p write the recprocal of the slope you found in part 1

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Answer:

The answer in the procedure

Step-by-step explanation:

we know that

If two lines are perpendicular, then the product of their slopes is equal to -1

so

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

we have

[tex]y=3x+5[/tex] -----> equation of line q

the slope of line q is

[tex]m1=3[/tex]

Find the slope of line p

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

[tex]3*m2=-1[/tex]

[tex]m2=-1/3[/tex] ----> slope of line p

Find the equation of the line p

The equation into point slope form is equal to

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

we have

[tex]point(6,-5)[/tex]

[tex]m=-1/3[/tex]

substitute

[tex]y+5=-(1/3)(x-6)[/tex]

[tex]y=-(1/3)x+2-5[/tex]

[tex]y=-(1/3)x-3[/tex] ----> equation of the line p

see the attached figure to better understand the problem

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