If a point P on the line 3x + 5y = 15 is equi | vedclass

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(B) Let the point $P(h, k)$ lie on the line $3x + 5y = 15$. Since $P$ is equidistant from the coordinate axes,we have $|h| = |k|$,which implies $h = k

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either in first or in second quadrant

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(B) Let the point $P(h, k)$ lie on the line $3x + 5y = 15$. Since $P$ is equidistant from the coordinate axes,we have $|h| = |k|$,which implies $h = k$ or $h = -k$. Case $1$: If $h = k$,then $3h + 5h = 15$ $\Rightarrow 8h = 15$ $\Rightarrow h = \frac{15}{8}$. Thus,$P = (\frac{15}{8}, \frac{15}{8})$,which lies in the first quadrant. Case $2$: If $h = -k$,then $3h + 5(-h) = 15$ $\Rightarrow -2h = 15$ $\Rightarrow h = -\frac{15}{2}$. Thus,$k = \frac{15}{2}$,and $P = (-\frac{15}{2}, \frac{15}{2})$,which lies in the second quadrant. Therefore,$P$ lies either in the first or in the second quadrant.

If a point $P$ on the line $3x + 5y = 15$ is equidistant from the coordinate axes,then $P$ lies

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