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

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

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]

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:

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

which can be written as

$\displaystyle \sqrt {a^{ - 2} + b^{ - 2} } = \frac{1}{p}$.

Now square both sides.

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

1 Attachment(s)

Re: Co-ordinate Geometry Q

Please refer to the sketch:

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):

$\displaystyle cp+dp=ab$

$\displaystyle p(c+d)=ab$

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

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

By Pythagoras, we have:

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

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