Question

Write an equation of the line through the point (5, 3) that is perpendicular to the line –3x+7y=5.

Answer 1

equation: y = -7/3x + 14 2/3

Explanation:

–3x+7y=5

y = 3/7 x + 5

slope of perpendicular line: - 7/3

y = mx + b for (5 , 3)

b = y - mx = 3 - ((-7/3) * 5) = 3 + 35/3 = 44/3 = 14 2/3

equation: y = -7/3x + 14 2/3

Answer 2

7x + 3y = 44

Explanation:

The equation of a line in slope- intercept form is

y = mx + c (m is the slope and c the y- intercept )

Rearrange - 3x + 7y = 5 into this form by adding 3x to both sides

7y = 3x + 5 ( divide all terms by 7 )

y =
`(3)/(7)` x +
`(5)/(7)` ← in slope- intercept form

with slope m =
`(3)/(7)`

Given a line with slope m then the slope of a line perpendicular to it is

`m_(perpendicular)` = -
`(1)/(m)` = -
`(1)/((3)/(7) )` = -
`(7)/(3)` , thus

y = -
`(7)/(3)` x + c ← is the partial equation

To find c substitute (5, 3) into the partial equation

3 = -
`(35)/(3)` + c ⇒ c = 3 +
`(35)/(3)` =
`(44)/(3)`

y = -
`(7)/(3)` x +
`(44)/(3)` ← in slope- intercept form

Multiply through by 3

3y = - 7x + 44 ( add 7x to both sides )

7x + 3y = 44 ← in standard form