SOLUTION: P(-3,1), Q(2,5) and R(7,a) are three points in the cartesian plane. Find the values of ''a'' in each case if P,Q and R are collinear PQR=90 degrees PR=QR

Question 848023: P(-3,1), Q(2,5) and R(7,a) are three points in the cartesian plane. Find the values of ''a'' in each case if
P,Q and R are collinear
PQR=90 degrees
PR=QR
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!

First find equation of line PQ.
Slope ;
y=mx+b
b=y-mx
Use the point P or Q, find b value.
;
Equation for PQ is .
-
R is collinear: Use x=7 to find "a".--------

.
Point R(7, 42/5) Collinear with P and Q.

PQR to be 90 degree---
That is perpendicular to eachother, PQ and QR.
The line QR would have slope .
The line for QR in this way, which will contain Q, will be y=-(5/4)x+b; here, this b is going to be different from the one for PQ.
b=y+(5/4)x
and using point Q(2, 5),
.
Line QR is .
-
What then is R? What is y when x=7?

Point R is (7, -5/4).

PR=QR-------
Use the distance formula for PR and QR. The variable will be a.
-

Equating PR=QR and squaring both sides gives
.
-...
You continue this to find a?
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