Question

Reduce the line $2x-3y+5=0$,
(a) In slope- intercept form and hence find slope & Y- intercept.
(b) In intercept form and hence find intercepts on the axes.
(c)In normal form and hence find perpendicular distance from the origin and angle made by the perpendicular with the positive X- axis.

Answer 1

(a)slope intercept form is y=mx

$y=\frac{2x}{3}+\frac{5}{3}$

therefore $slope=\frac{2}{3}$,$y-intercept=\frac{5}{3}$

(b)y-intercept=$\frac{5}{3}$,x-intercept=$\frac{-5}{2}$

there fore intercept form is $\frac{x}{\frac{-5}{2}}+\frac{y}{\frac{5}{3}}=1$

(c)for converting the given line into normal form, divide the equation ax+by+c=0 by √(a²+b²).

equation of line in normal form is $\frac{2x}{\sqrt{1}3}-\frac{3y}{\sqrt{1}3}=-\frac{5}{\sqrt{1}3}$

there fore perpendicular distance is $\frac{-5}{\sqrt{1}3}$