Question

The intercepts of a straight line on the axes of coordinates are a and b. If p is the length of the perpendicular drawn from the origin to the line, write the value of p in terms of a and b....




please do it fast..​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{The intercepts of a straight line on the axes of coordinates are a and b}$

$\textsf{p is the length of the perpendicular drawn from the origin to the line}$

$\underline{\textsf{To find:}}$

$\textsf{The value of p in terms of a and b}$

$\underline{\textsf{Solution:}}$

$\textsf{The equation of the given line}$

$\textsf{can be written in intercept form as}$

$\mathsf{\dfrac{x}{a}+\dfrac{y}{b}=1}$.....(1)

$\textsf{The same line can be written in normal form as}$

$\mathsf{x\,cos\alpha+y\,sin\alpha=p}$

$\textsf{where}\,\mathsf{\alpha}\,\textsf{is angle inclined by the perpendicular to x axis}$

$\textsf{Divide bothsides by p}$

$\mathsf{\dfrac{x\,cos\alpha}{p}+\dfrac{y\,sin\alpha}{p}=1}$

$\mathsf{\dfrac{x}{\dfrac{p}{cos\alpha}}+\dfrac{y}{\dfrac{p}{sin\alpha}}=1}$......(2)

$\textsf{Comparing (1) and (2), we get}$

$\mathsf{\dfrac{p}{cos\alpha}=a\;\;\;\&\;\;\;\dfrac{p}{sin\alpha}=b}$

$\mathsf{\dfrac{p}{a}=cos\alpha\;\;\;\&\;\;\;\dfrac{p}{b}=sin\alpha}$

$\textsf{We know that,}\,\boxed{\mathsf{cos^2\alpha+sin^2\alpha=1}}$

$\implies\mathsf{\dfrac{p^2}{a^2}+\dfrac{p^2}{b^2}=1}$

$\implies\mathsf{p^2(\dfrac{1}{a^2}+\dfrac{1}{b^2})=1}$

$\implies\mathsf{p^2(\dfrac{b^2+a^2}{a^2b^2})=1}$

$\implies\mathsf{p^2=\dfrac{a^2b^2}{a^2+b^2}}$

$\implies\boxed{\mathsf{p=\displaystyle\sqrt{\dfrac{a^2b^2}{a^2+b^2}}}}$

$\underline{\textsf{Answer:}}$

$\textsf{The value of p is}$

$\mathsf{\displaystyle\sqrt{\dfrac{a^2b^2}{a^2+b^2}}}}$