Question

Find the length of the perpendicular from the point $(x^1,y^1)$ to the straight line Ax + By + C = 0, the axes being inclined at an angle $\omega$, and the equation being written such that C is a negative quantity.

• A. $\frac{A{x}^1+B{y}^1+C}{\sqrt{A^2+B^2}}.cos\omega$
• B. $\frac{A{x}^1+B{y}^1+C}{\sqrt{A^2+B^2}}.sin\omega$
• C. $\frac{A{x}^1+B{y}^1+C}{\sqrt{A^2+B^2-2ABsin\omega }}.cos\omega$
• D. $\frac{A{x}^1+B{y}^1+C}{\sqrt{A^2+B^2-2ABcos\omega }}.sin\omega$

Answer 1

The correct option is D

$\frac{A{x}^1+B{y}^1+C}{\sqrt{A^2+B^2-2ABcos\omega }}.sin\omega$

Let the given straight line meet the axis in L and M. So that $OL=-\frac{C}{A}andOM=\frac{-C}{B}$
Let P be the given point ($x^1,y^1$). Draw the perpendicular PQ, PR and PS on the given line and two axes.
Taking O and P on opposite sides of the given line, we then have
$\Delta LPM+\Delta MOL=\Delta OLP+\Delta OPM$
i.e. PQ.LM + OL. OM sin $\omega$ = OL.PR + OM. PS - - - (1)
Draw PU and PV parallel to the axes of y and x, such that PU=$y^1$ and PV = $x^1$
Hence, PR = PU sin PUR =$y^{1}sin\omega$
and PS = PV sin PVS = $x^{1}sin\omega$
Also, $LM=\sqrt{O{L}^2+O{M}^2-2OL.OMcos\omega }=\sqrt{\frac{C^2}{A^2}+\frac{C^2}{B^2}-2\frac{C^2}{AB}cos\omega }=-C\sqrt{\frac{1}{A^2}+\frac{1}{B^2}\frac{-2cos\omega }{AB}}$
Since, C is a negative quantity
On substituting these values in (1), we have
$PQ\times (-C)\times \sqrt{\frac{1}{A^2}+\frac{1}{B^2}-\frac{2cos\omega }{AB}}+\frac{C^2}{AB}sin\omega =-\frac{C}{A}{y}^{1}sin\omega -\frac{C}{B}{x}^{1}sin\omega$
On simplifying,
$PQ=\frac{A{x}^1+B{y}^1+C}{\sqrt{A^2+B^2-2ABcos\omega }}.sin\omega$