# Co-ordinate Geometry Q

• Jan 5th 2013, 03:06 AM
HelenMc9
Co-ordinate Geometry Q
A question I'm stuck on, I'm never quite sure how to start off questions that require you to 'prove' things ... I don't have a very mathematical brain I think! :/ so here it is:

if p is the length if the perpendicular from the origin to the line x/a + y/b = 1, prove that 1/p² = 1/a² + 1/b².

I assume I'll have to use the perpendicular distance formula somehow, but I don't really know what to do ...
• Jan 5th 2013, 03:33 AM
Asharma
Re: Co-ordinate Geometry Q
We know that the length of the perpendicular from the point (a', b') to the line ax+by+c=0 is (aa'+bb'+c)/[sqrt(a^2+b^2)]
In our case a'=0, b'=0
and the line is x/a+y/b=1
thus substituting in the formula we get
(0+0-1)/[sqrt[(1/a^2)+(1/b^2)]= p
-1/sqrt[1/a^2+1/b^2] =p
resiprocating both the sides
sqrt[1/a^2+1/b^2]=1/p
square both the sides we get
1/a^2+1/b^2=1/p^2
• Jan 5th 2013, 03:52 AM
ibdutt
Re: Co-ordinate Geometry Q
The following may further help in understanding
From the step
(-1)/(√(1/a^2 + 1/b^2 ) )=p By squaring both the sides we get

1/((a^2+ b^2)/(a^2 b^2 ) )=p^2
1/p^2 =(a^2+ b^2)/(a^2 b^2 )
1/p^2 = a^2/(a^2 b^2 )+ ( b^2)/(a^2 b^2 )

1/p^2 = 1/b^2 + ( 1)/a^2
• Jan 5th 2013, 04:30 AM
HelenMc9
Re: Co-ordinate Geometry Q
Ok I understand everything up to here but how to do you get this when you square it?:

1/((a^2+ b^2)/(a^2 b^2 ) )=p^2 ([/QUOTE]
• Jan 5th 2013, 07:33 AM
Plato
Re: Co-ordinate Geometry Q
Quote:

Originally Posted by HelenMc9
Ok I understand everything up to here but how to do you get this when you square it?:
1/((a^2+ b^2)/(a^2 b^2 ) )=p^2

Using the distance formula for a point to a line we get:
$\frac{1}{{\sqrt {a^{ - 2} + b^{ - 2} } }} = p$
which can be written as
$\sqrt {a^{ - 2} + b^{ - 2} } = \frac{1}{p}$.

Now square both sides.
• Jan 5th 2013, 09:18 PM
ibdutt
Re: Co-ordinate Geometry Q
The denominator was under the square root. It was Sqr root √(1/a^2 + 1/b^2 ). I hope it is clear now
• Jan 5th 2013, 10:00 PM
MarkFL
Re: Co-ordinate Geometry Q

Attachment 26486

The sum of the areas of the two smaller triangles is equal to the area of the largest triangle, giving us (after multiplying by 2):

$cp+dp=ab$

$p(c+d)=ab$

$\frac{1}{p}=\frac{c+d}{ab}$

$\frac{1}{p^2}=\frac{(c+d)^2}{a^2b^2}$

By Pythagoras, we have:

$(c+d)^2=b^2+a^2$ and so:

$\frac{1}{p^2}=\frac{b^2+a^2}{a^2b^2}=\frac{1}{a^2} +\frac{1}{b^2}$