find the equation of the line perpendicular to the line 4x+3y=9 that passes through the point (-1,2)

Question

find the equation of the line perpendicular to the line 4x+3y=9 that passes through the point (-1,2)

Answer 1

To find the equation of a line perpendicular to the line 4x + 3y = 9 that passes through the point (-1, 2), you can follow these steps:

1. First, find the slope of the given line 4x + 3y = 9 by rearranging it into slope-intercept form (y = mx + b), where m is the slope:

4x + 3y = 9

3y = -4x + 9

y = (-4/3)x + 3

So, the slope of the given line is -4/3.

2. The slope of a line perpendicular to this one will be the negative reciprocal of -4/3, which is 3/4.

3. Now that you have the slope (m = 3/4) and a point on the line (-1, 2), you can use the point-slope form of a linear equation to find the equation of the perpendicular line:

y - y₁ = m(x - x₁)

Plug in the values:

y - 2 = (3/4)(x - (-1))

4. Simplify and solve for y:

y - 2 = (3/4)(x + 1)

Distribute (3/4) to both terms in the parentheses:

y - 2 = (3/4)x + 3/4

5. Add 2 to both sides of the equation:

y = (3/4)x + 3/4 + 2

y = (3/4)x + 11/4

So, the equation of the line perpendicular to 4x + 3y = 9 that passes through the point (-1, 2) is:

y = (3/4)x + 11/4

Answer 2

This is definitely not a Calculus problem.

Determine the slope of 4x + 3y = 9, name it "m"

Determine -1/m

Calclulate (y - 2) = (-1/m)(x + 1)