Question

Write the equation of the line passing through (-3, 4) and normal to -2x + 7y = -3. (Note: normal means perpendicular)

Answer 1

To find the equation of the line that is normal to -2x + 7y = -3, we need to find the slope of the given line and the negative reciprocal of the slope. Using the point-slope form of a line and the given point (-3, 4), we can find the equation of the normal line.

Step-by-step explanation:

To find the equation of a line that is normal to -2x + 7y = -3, we need to find the slope of the given line and then find the negative reciprocal of the slope to get the slope of the normal line.
First, rearrange the given equation in slope-intercept form (y = mx + b) by solving for y.
-2x + 7y = -3
7y = 2x - 3
y = (2/7)x - 3/7

The slope of the given line is 2/7. Since the normal line is perpendicular to the given line, the slope of the normal line is the negative reciprocal of 2/7, which is -7/2.

Now, we have the slope (-7/2) and a point (-3, 4) that the normal line passes through. We can use the point-slope form of a line to find the equation of the normal line:
y - y1 = m(x - x1)
y - 4 = (-7/2)(x - (-3))
y - 4 = (-7/2)(x + 3)
y - 4 = (-7/2)x - 21/2
y = (-7/2)x - 13/2

Answer 2

The answer to your question is y = -7/2x -13/2

Step-by-step explanation:

Point = (-3, 4)

Line -2x + 7y = -3

Process

1.- Solve equation for y to find the slope of the perpendicular line

7y = 2x - 3

y = 2/7 x - 3/7

slope = 2/7

Slope of the perpendicular line = -7/2

2.- Find the equation of the new line

y - y1 = m(x - x1)

y - 4 = -7/2(x + 3)

y - 4 = -7/2x - 21/2

y = -7/2x - 21/2 + 4

y = -7/2x -21/2 + 8/2

3.- Result

y = -7/2x -13/2