A line passing through (5, –6) and (–9, y) is perpendicular to a line with slope . Find the value of y.?

Question

I am stuck on this if someone can help me with it I thank you.

slope was 14/5 sorry

Answer 1

You failed to include the value of the slope when you copied the question.

But in general, you would take the slope (m). A perpendicular line would have a slope equal to the negative reciprocal (-1/m).

Next you have two points (5, -6) and (-9, y). The slope between two points is the rise (difference of y terms) divided by the run (difference of x terms).

rise = y - (-6) = y + 6
run = -9 - (-5) = -9 + 5 = -4

So the slope would be:
y + 6
-------
.. -4

And you would equate this to your negative reciprocal slope:
y + 6 ... -1
------- = ---
.. -4 ..... m

Cross multiplying:
(y + 6)m = 4

Now solve for y:
(y + 6) = 4/m
y = 4/m - 6

Again I don't know your slope, but if you have a value, just plug it in and you will have the value of y for the second point.

For example, if the slope of the original line was 2, then the y coordinate of the point would be:
y = 4/(2) - 6
y = 2 - 6
y = -4

Or if the slope was 1/4, then the y coordinate would be:
y = 4/(1/4) - 6
y = 16 - 6
y = 10

But again, it depends on your value of m.

Answer 2

The line passing through (5,-6) and (-9,y) has as its
slope {y(2)-y(1)}/ {x(2)-x(1)}.
This is y-(-6)/-9-5, which simplifies to (y+6)/-14.
This particular line is also perpendicular to a line whose slope is 14/5.
Because they are perpendicular to each other, their
slopes are negative reciprocals of one another.
This means that (y+6)/-14=-5/14
-(y+6)/14= -5/14
-(y+6)=-5
-y-6= -5
-y= -5+6
-y=1
y= -1