Question

Reduce the line $2x-3y+5=0$,
(i) in slope -intercept form and hence find slope and y-intercept
(ii) in intercept form and hence find intercepts on the axes
(iii) in normal form and hence find perpendicular distance from the origin and angle made by the perpendicular with the positive x-axis

Answer 1

Given, $2x-3y+5=0$

(i) Rewriting as $y=\frac{2x}{3}+\frac{5}{3}$ & Comparing with $y=mx+c$

slope = $m=\frac{2}{3}$, y-intercept = $c=5/3$

(ii) Rewriting as $\frac{-2}{5}x+\frac{3}{5}y=1$ & comparing with $\frac{x}{a}+\frac{y}{b}=1$

x-intercept = $-5/2$ y-intercept = $5/3$

(iii) Let equation in Normal form be $xcos\alpha +ysin\alpha =p$

Then, $\frac{cos\alpha }{2}=\frac{sin\alpha }{-3}=\frac{-p}{5}=\frac{\sqrt{co{s}^2\alpha +si{n}^2\alpha }}{\sqrt{2^2+3^2}}=\frac{1}{\sqrt{13}}$

$\Rightarrow cos\alpha =\frac{-2}{\sqrt{13}},sin\alpha =\frac{+3}{\sqrt{13}},p=\frac{+5}{\sqrt{13}}$

Normal equation : $\frac{-2}{\sqrt{13}}\times \frac{+3}{\sqrt{13}}y=\frac{5}{\sqrt{13}}$

Perpendicular from origin = $p=\frac{5}{\sqrt{13}}$, angle with x-axes = $\alpha =ta{n}^{-1}\left(\frac{-3}{2}\right)$