Find the equation of the line perpendicular to -3x-5y=7 and having x- intercept of -8.

Question

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Answer 1

For ax + by = c

the slope is found by m = -a/b

 

-3x - 5y = 7

slope:  m = -(-3)/(-5) = 3/(-5) = -3/5

 

Slope of a perpendicular line is the 

negative reciprocal:  m_{⊥ = -1/m}

The perpendicular slope is -1/(-3/5) = 5/3

 

If the x-intercept is -8 then the point

(-8, 0) is on the line.

 

Now you have slope 5/3 and a point (-8,0).

Use point slope form: y - y₁=m(x-x₁) with

(x₁, y₁) = (-8, 0) and m = 5/3

 

y - 0 = (5/3)(x-(-8))

y = (5/3)(x+8)

3y = 5(x+8)

3y = 5x + 40

-5x + 3y = 40

or

5x - 3y = -40 

 

Convention normally has the coefficient of x as

a positive number; however, both equations

are correct.