A point on the straight line 3x + 5y = 15 whi | vedclass

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(D) Let the point be $P(t, y)$. Since the point lies on the line $3x + 5y = 15$,we have $3t + 5y = 15$,which gives $y = \frac{15 - 3t}{5}$.Since the p

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$1^{st}$ and $2^{nd}$ quadrants

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(D) Let the point be $P(t, y)$. Since the point lies on the line $3x + 5y = 15$,we have $3t + 5y = 15$,which gives $y = \frac{15 - 3t}{5}$.Since the point is equidistant from the coordinate axes,$|x| = |y|$,so $|t| = |\frac{15 - 3t}{5}|$.This implies $\frac{15 - 3t}{5} = t$ or $\frac{15 - 3t}{5} = -t$.Case $1$: $15 - 3t = 5t$ $\Rightarrow 8t = 15$ $\Rightarrow t = \frac{15}{8}$. Then $y = \frac{15}{8}$. Point $P(\frac{15}{8}, \frac{15}{8})$ lies in the $1^{st}$ quadrant.Case $2$: $15 - 3t = -5t$ $\Rightarrow 2t = -15$ $\Rightarrow t = -\frac{15}{2}$. Then $y = \frac{15}{2}$. Point $P(-\frac{15}{2}, \frac{15}{2})$ lies in the $2^{nd}$ quadrant.Thus,the points lie in the $1^{st}$ and $2^{nd}$ quadrants.

$A$ point on the straight line $3x + 5y = 15$ which is equidistant from the coordinate axes will lie only in

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