SOLUTION: P and Q are the points of intersection of the line x/a +y/b = 1 ( a > 0 , b > 0),with the x and y axes respectively. The distance PQ is 10 and the gradient of PQ is -2. Find the

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Question 1178030: P and Q are the points of intersection of the line x/a +y/b = 1 ( a > 0 ,
b > 0),with the x and y axes respectively. The distance PQ is 10 and the
gradient of PQ is -2. Find the value of a and of b.
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing it for you, I'll change the numbers and do one exactly
step-by-step like yours.  Here's the one I'll do:

P and Q are the points of intersection of the line x/a +y/b = 1
(a > 0, b > 0), with the x and y axes respectively. The distance PQ is 20
and the gradient of PQ is -3. Find the value of a and of b.



We use the gradient formula:

%280-b%29%2F%28a-0%29=-3
-b%2Fa=-3
b%2Fa=3
b=3a

We use the distance formula:

sqrt%28%28a-0%29%5E2%2B%280-b%29%5E2%29=20
sqrt%28a%5E2%2Bb%5E2%29=20
%28sqrt%28a%5E2%2Bb%5E2%29%29%5E2=20%5E2
a%5E2%2Bb%5E2=400
Substitute 3a for b
a%5E2%2B%283a%29%5E2=400
a%5E2%2B9a%5E2=400
10a%5E2=400
a%5E2=40
a=sqrt%2840%29  <--positive square root since a > 0
a=sqrt%284%2A10%29
a=2sqrt%2810%29

b=3a
b=3%2A2sqrt%2810%29
b=6sqrt%2810%29

Now do yours the exact same way.

Edwin