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

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(A) Let $P$ be a point $(x, y)$ that is equidistant from the coordinate axes. This implies $|x| = |y|$,so the point must lie on the lines $y = x$ or $

Vedclass · Accepted Answer

either $1^{	ext{st}}$ quadrant or $2^{	ext{nd}}$ quadrant

Vedclass · Answer

(A) Let $P$ be a point $(x, y)$ that is equidistant from the coordinate axes. This implies $|x| = |y|$,so the point must lie on the lines $y = x$ or $y = -x$. Case $1$: Intersection of $3x + 5y = 15$ and $y = x$. Substituting $y = x$ into the equation: $3x + 5x = 15$ $\Rightarrow 8x = 15$ $\Rightarrow x = \frac{15}{8}$. Thus,$y = \frac{15}{8}$. The point is $(\frac{15}{8}, \frac{15}{8})$,which lies in the $1^{	ext{st}}$ quadrant. Case $2$: Intersection of $3x + 5y = 15$ and $y = -x$. Substituting $y = -x$ into the equation: $3x + 5(-x) = 15$ $\Rightarrow 3x - 5x = 15$ $\Rightarrow -2x = 15$ $\Rightarrow x = -\frac{15}{2}$. Thus,$y = -(-\frac{15}{2}) = \frac{15}{2}$. The point is $(-\frac{15}{2}, \frac{15}{2})$,which lies in the $2^{	ext{nd}}$ quadrant. Therefore,the point lies in either the $1^{	ext{st}}$ or $2^{	ext{nd}}$ quadrant.

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

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