Question

The equation of a line is $ax+by-c=0wherea,b,c>0$

Answer 1

A) The equation of the given line can be written as $ax+by=c$
$\Rightarrow \frac{x}{\left(c/a\right)}+\frac{y}{\left(c/b\right)}=1$
Gives x intercept =$\frac{c}{a}$ and y intercept =$\frac{c}{b}$
then sum of the intercepts on coordinate axes
$=\frac{c}{a}+\frac{c}{b}=\frac{c(a+b)}{ab}(\therefore c>0)$
B) Length of the perpendicular from the origin on the line
$=\left|\frac{-c}{\sqrt{a^2+b^2}}\right|=\frac{c}{\sqrt{a^2+b^2}}(\because c>0)$
C) Length of the line segment $AB$ intercepted between the coordinates axes
$=\sqrt{{\left(\frac{c}{a}\right)}^2+{\left(\frac{c}{b}\right)}^2}=\frac{c\sqrt{a^2+b^2}}{ab}$
D) Area of the triangle formed by the line and the coordinates axes $=\frac{1}{2}\times \frac{c}{a}\times \frac{c}{b}=\frac{c^2}{2ab}$