Question

3. What is the equation of the line perpendicular to that passes through ? Write your answer in slope-intercept form. Show your work.
Answer:

Answer 1

I don't see the given line or the point in your question — can you paste them?

Meanwhile, here's the method and a worked example so you can apply it:

Method
1. Put the given line into slope-intercept form y = m1 x + b1 and read off its slope m1.
2. The slope of any line perpendicular to it is m2 = −1/m1 (except special cases below).
3. Use the point (x0,y0) the new line must pass through and the point-slope form:
y − y0 = m2 (x − x0).
4. Solve for y to get slope-intercept form y = m2 x + b, where b = y0 − m2 x0.

Special cases
- If the given line is vertical (x = c), its slope is infinite; a perpendicular line is horizontal: y = y0.
- If the given line is horizontal (y = k), its slope is 0; a perpendicular line is vertical: x = x0 (not expressible as y = mx + b).

Example
Given line 2x + 3y − 6 = 0 and point (4,1).
1. Solve for y: 3y = −2x + 6 → y = (−2/3)x + 2, so m1 = −2/3.
2. Perpendicular slope m2 = −1/m1 = −1/(−2/3) = 3/2.
3. Use point (4,1): y − 1 = (3/2)(x − 4).
4. Solve: y − 1 = (3/2)x − 6 → y = (3/2)x − 5.

So the perpendicular line through (4,1) is y = (3/2)x − 5.

If you send the specific line and point from your problem I will compute the exact answer.