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Q.
A straight line with slope $3$ intersects a straight line with slope $6$ at the point $(30, 40)$. Then the difference between the $y$-intercepts of the straight lines is
Equation of straight line with slope 3 and passing through the point $(30,40)$ is
$(y-40)=3(x-30)$
$\Rightarrow \, y-40=3 x-90$
$\Rightarrow \, 3 x-y=50$
$\Rightarrow \, \frac{x}{\left(\frac{50}{3}\right)}+\frac{y}{(-50)}=1\,\,\,\,\,\,\dots(i)$
and equation of straight line with slope 6 and passing through the point $(30,40)$ is
$ (y-40)=6(x-30) $
$\Rightarrow \, y-40=6 x-180 $
$\Rightarrow \, 6 x-y=140 $
$\Rightarrow \, \frac{x}{\left(\frac{70}{3}\right)}+\frac{y}{(-140)}=1\,\,\,\,\,\dots(ii)$
$\therefore $ The difference between the $y$ -intercepts of the straight lines $=-50-(-140)=140-50=90$