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

If you know how to do it, please help me out :)-example-1
User Deepak Rao
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1 Answer

1 vote

Answer:

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

User Arpita
by
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