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Two parallel lines lying in the
same quadrant make
intercepts a and bon x, y
axes respectively between
them, then the distance
between the lines is​

Answer 1

GIVEN :

Two parallel lines lying in the  same quadrant make  intercepts a and b on x, and  y  axes respectively, between  them.

TO FIND :

The distance  between the two parallel lines

SOLUTION :

Given that two parallel lines lying in the  same quadrant make  intercepts a and b on x, and  y  axes respectively, between  them.

From the given that we have ,

In ΔABC we get

$sin\theta=\frac{d}{a}$

And In ΔPQR we get

$cos\theta=\frac{d}{b}$

We know that $sin^2\theta+cos^2\theta=1$

By substituting the values we get,

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

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

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

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

Rewritting we get,

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

$d^2(a^2+b^2)=a^2b^2$

$d^2=\frac{a^2b^2}{a^2+b^2}$

$d=\frac{ab}{\sqrt{a^2+b^2}}$

∴ the distance between the parallel lines is $d=\frac{ab}{\sqrt{a^2+b^2}}$.