Write an equation of the line that passes through the point (1,5) and is (a) parallel to the line y=3x-5 and (b) perpendicular to the line y=3x-5

Answer 1

The line parallel to this one will have the same slope. So m=3 Then you can use the following equation.

(y-y1)=m(x-x1) Plug in the values of (x1,y1) using the point provided. (1,5)

(y-5)=3(x-1) Distribute

y-5=3x-3 Add 5 to both sides

y-5+5=3x-3+5 Simplify

y=3x+2

The line perpendicular to this one will have a slope equal to the negative reciprocal. So m=-1/3 Then you can use the following equation.

(y-y1)=m(x-x1) Plug in the values of (x1,y1) using the point provided. (1,5)

(y-5)=-1/3(x-1) Distribute

y-5=-1/3x+1/3 Add 5 to both sides

y-5+5=-1/3x+1/3+5 Simplify

y=-1/3x+16/3

Answer 2

(a) the slope of y=3x-5 is the coefficient of the x term, or 3

Use the point slope formula: y-k=m(x-h) (h,k) is the point (1,5) and m=the slope=3

A line going through (1,5) with slope of 3 is y-5=3(x-1) or y=3x+2

Check that equation by plugging in (1,5) to get 5=3(1)+2

(b) Perpendicular means the slope is the negative inverse of 3 or -1/3

Then the line through (1,5) with slope = -1/3 is y-5=-(1/3)(x-1) or 3y+x=16

check that by plugging in (1,5) to get 3(5)+1=16