A point on the straight line $3 x+5 y=15$ which is equidistant from the coordinate axes will lie in
A point on the straight line $3 x+5 y=15$ which is equidistant from the coordinate axes will lie in
either $1^{\text {st }}$ quadrant or $2^{\text {nd }}$ quadrant
$4^{\text {th }}$ quadrant only
$3^{\text {rd }}$ quadrant only
either in the $3^{\text {rd }}$ or in the $4^{\text {th }}$ quadrant
Solution
Let $P$ be the point which is equidistant from the coordinate axis.
$\therefore \quad P$ lies on the line $y=x$ or $y=-x$.
Let us find intersecting point of $3 x+5 y=15$ and $y=x$ for this,
$3 x+5 x=15 \Rightarrow 8 x=15 \Rightarrow x=\frac{15}{8}$
So, $y=x=\frac{15}{8}$
$\therefore$ Intersection point of $\left(\frac{15}{8}, \frac{15}{8}\right)$, which is in first quadrant.
Now, let us find the intersecting point of $3 x+5 y=15$ and $y=-x$ for this
$\Rightarrow 3 x-5 x=15 \Rightarrow-2 x=15 \Rightarrow x=\frac{-15}{2}$
So, $y=-x=\frac{15}{2}$
$\therefore$ Intersecting point is $\left(-\frac{15}{2}, \frac{15}{2}\right)$, which is in 2nd quadrant.