Question

If you know how to do it, please help me out :)

Answer 1

Answer in explanation.

Explanation:

Lets take

`y = (3)/(4) x - 5`

as L1.

Take L2 the line perpendicular to L1.

Standard form of equation of line:

y=mx+c, where m = slope and c = y-intercept.

Since L1 and L2 are perpendicular,

mL1 x mL2 = -1

Substitute mL1 into the equation,

3/4 x mL2 = -1

mL2 = -1 ÷ 3/4

mL2 = -4/3

L2 : y = mx+ c

Substitute y = 6, x = 8 and m = -4/3 into the equation,

`6 = - (4)/(3) (8) + c \\ 6 = - (32)/(3) + c \\ c = 6 + (32)/(3) \\ = 16 (2)/(3)`

therefore L2:

`y = - (4)/(3)x + 16 (2)/(3)`

Lets take L3 as the line parallel to L1.

Since L3 and L1 are parallel,

mL3 = mL1 = 3/4

equation of line: y = mx+c

substitute y = 6, x = 8 and m = 3/4 into equation.

`6 = ( (3)/(4) )(8) + c \\ 6 = 6 + c \\ c = 6 - 6 \\ = 0`

therefore L3:

`y = (3)/(4) x`