Question

Find the equation for the line passing through the point (-1, 1) and perpendicular to the line whose equation is -2x-3y-5=0

Answer 1

y =
`(3)/(2)`x +
`(5)/(2)`

Explanation:

1. Put the given equation in Y = mx + b form

-2x-3y-5=0

-2x-5=3y

-2/3x - 5/3=y

2. Since We know that the slope of the given line is perpendicular, therefore the slope of the new line is the negative recipricol

m = 3/2 (slope of new line)

3. Find your b value by subbing in the points (-1,1)

y=mx + b

1= 3/2 (-1) + b

1= -3/2 +b

1+ 3/2 = b

5/2= b

4. write your equation

y = 3/2 x + 5/2

Answer 2

The answer to your question is y = 3/2x + 5/2

Explanation:

Data

P (-1, 1)

⊥ -2x - 3y - 5 = 0

Process

1.- Find the slope of the original line

-2x - 3y - 5 = 0

-3y = 2x + 5

y = -2/3x - 5/3

Slope = -2/3

2.- Find the slope of the new line

Slope = 3/2

3.- Find the equation of the line

y - y1 = m(x - x1)

-Substitution

y - 1 = 3/2(x + 1)

-Simplification

y - 1 = 3/2x + 3/2

y = 3/2x + 3/2 + 1

-Result

y = 3/2x + 5/2