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

Algebra ->  Length-and-distance -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!

First find equation of line PQ.
Slope m=%285-1%29%2F%282-%28-3%29%29=4%2F5;
y=mx+b
b=y-mx
Use the point P or Q, find b value.
b=5-%284%2F5%292=5-8%2F5=25%2F5-8%2F5=17%2F5;
Equation for PQ is highlight%28y=%284%2F5%29x%2B17%2F5%29.
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R is collinear: Use x=7 to find "a".--------
y=%284%2F5%29%2A7%2B17%2F5
y=28%2F5%2B17%2F5=highlight%2842%2F5%29.
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 m=-5%2F4.
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),
b=5%2B%285%2F4%292=5%2B10%2F4=20%2F4%2B10%2F4=30%2F4=15%2F2.
Line QR is highlight%28y=-%285%2F4%29x%2B15%2F2%29.
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What then is R? What is y when x=7?
y=-%285%2F4%29%2A7%2B15%2F2
y=-35%2F4%2B15%2F2=-35%2F4%2B30%2F4=-5%2F4
Point R is (7, -5/4).


PR=QR-------
Use the distance formula for PR and QR. The variable will be a.
-
PR=sqrt%28%287-%28-3%29%29%5E2%2B%28a-1%29%5E2%29
sqrt%28100%2Ba%5E2-2a%2B1%29
sqrt%28a%5E2-2a%2B101%29

QR=sqrt%28%287-2%29%5E2%2B%28a-5%29%5E2%29
sqrt%2825%2Ba%5E2-10a%2B25%29
sqrt%28a%5E2-10a%2B50%29

Equating PR=QR and squaring both sides gives
highlight_green%28a%5E2-2a%2B101=a%5E2-10a%2B50%29.
-...
You continue this to find a?