Suppose the line 2x-3y=5 bisects a line PQ at right angles. If P is the point (-1,2), find the coordnates of Q. Please help me with this!!?

slope of PQ = -3/2
y-2=(-3/2)(x+1)
2y-4=-3x-3
3x+2y = 1
2x-3y = 5
9x+6y=3
4x-6y=10
13x=13
x = 1
y = -1
Intersection point (1,-1) is midpoint of PQ
Q = (x,y)
-1+x = 2
2+y = -2
Q is (3,-4)

It's a lot easier if you rewrite the equation of the given line in slope-intercept form:
y = mx + b
where m and b are numbers. In this form, m is the slope of the line and (0,b) is the y-intercept.

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

So the slope of our given line is 2/3. Two lines are perpendicular (i.e. intersect at right angles) if their slopes are negative reciprocals of each other. So if one line has slope a/b, the perpendicular line will have slope -b/a.

Our given line has slope 2/3, as we proved. So line PQ must have slope -3/2.

We know PQ must also have the form
y = mx + b
We know m (the slope) and have an (x,y) given to us as point P (-1,2). Let's solve for b.
y = mx + b
2 = (-3/2)(-1) + b
2 = 3/2 + b
1/2 = b

So line PQ in slope-intercept form is
y = (-3/2)x + 1/2

Where does PQ intersect our original line? Where the same (x,y) satisfies both equations.
y(given) = (2/3)x - 5/3
y(PQ) = (-3/2)x + 1/2

These should be set equal, then solve for x.
(2/3)x - 5/3 = (-3/2)x + 1/2
Multiply both sides by 6 to clear the fractions.
4x - 10 = -9x + 3
13x - 10 = 3
13x = 13
x = 1

Now solve for y using either equation. Let's use the one for PQ:
y(PQ) = (-3/2)x + 1/2
y(PQ) = (-3/2)(1) + 1/2
y(PQ) = -1

So the two lines intersect at the (x,y) of (1,-1). In fact we know they BISECT there, so (1,-1) is halfway between P and Q. That means 1 is midway between the x-values of P and Q, and -1 is midway between the y-values of P and Q.

Average of x-values from P and Q = 1
Average of y-values from P and Q = -1

So the average of -1 and Q's x-value is 1. That means Q's x-value is 3.
And the average of 2 and Q's y-value is -2. That means Q's y-value is -6.

So Q is (3, -6). Hope that makes sense.

@Mizoo

I'm sorry, but this is extremely confusing. Sorry I can't help ya'.