Question 1178030
<pre>
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:
</pre>
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.<pre>

{{{drawing(750/3,600,-2.5,7,-2.5,20,graph(750/3,600,-2.5,7,-2.5,20),
line(0,18.97366596,6.32455532), locate(.1,19.4,"(0,b)"),
locate(4.5,.7,"(a,0)"),locate(3.2,10,20))}}}

We use the gradient formula:

{{{(0-b)/(a-0)=-3}}}
{{{-b/a=-3}}}
{{{b/a=3}}}
{{{b=3a}}}

We use the distance formula:

{{{sqrt((a-0)^2+(0-b)^2)=20}}}
{{{sqrt(a^2+b^2)=20}}}
{{{(sqrt(a^2+b^2))^2=20^2}}}
{{{a^2+b^2=400}}}
Substitute 3a for b
{{{a^2+(3a)^2=400}}}
{{{a^2+9a^2=400}}}
{{{10a^2=400}}}
{{{a^2=40}}}
{{{a=sqrt(40)}}}  <--positive square root since a > 0
{{{a=sqrt(4*10)}}}
{{{a=2sqrt(10)}}}

{{{b=3a}}}
{{{b=3*2sqrt(10)}}}
{{{b=6sqrt(10)}}}

Now do yours the exact same way.

Edwin</pre>