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# Help, write an equation... ?

I asked this earlier but wrote the question wrong. Write an equation in standard form for a line that is (a) parallel (b) perpendicular to the line with an equation of y = 3x - 5 that passes through the point (8, 5).

### 3 Answers

- Φ² = Φ+1Lv 72 months ago
(a) y = 3x + 5 - 3×8 which is y = 3x - 19

(b) y = (-1/3)x + 5 + 8/3 which is y = -⅓x + ²³∕₃

Source(s): The line parallel to y=mx+b through a given point (x₀,y₀) is y = mx + y₀ - mx₀, while the line perpendicular to y=mx+b through a given point (x₀,y₀) is y = (-1/m)x + y₀ + x₀/m.- Login to reply the answers

- King LeoLv 72 months ago
Equation of a straight line with a slope m and given point( a, b )

y - b = m( x - a )

y = 3x - 5

slope: m₁ = 3

parallel lines: have same slope ===> m₁ = m₂

perpendicular lines: product of their slopes is negative one ===> m₁m₂ = -1

( a )Parallel:

m₁ = m₂ = 3

( a, b ) = ( 8, 5 )

y - 6 = 3( x - 8 )

y - 6 = 3x - 24

y = 3x - 18

——————

Perpendicular:

m₁m₂ = -1

m₂ = -⅓

( a, b ) = ( 8, 5 )

y - 6 = -⅓( x - 8 )

y - 6 = -⅓x - 8/3

y = -⅓x + 10/3

——————

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- King LeoLv 72 months agoReport
you are welcome

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- TasmLv 62 months ago
y = 3x - 5

parallel just change the last number

y = 3x - 6

perpendicular just reverse the sign of the x value

y = -3x - 5

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