# A point on the straight line, $3x+5y=15$ which is equidistant from the coordinate axes will lie only in :  Option 1)$1^{st}$ quadrantOption 2)$1^{st},2^{nd}and \; 4^{th}$ quadrantsOption 3)$4^{th}$ quadrant  Option 4)$1^{st} \; and\; 2^{nd}$ quadrants

Let the point will be $\\(h,k)\\since\;point\;is\;equidistance\;from\;the\;axes\\h=k$

$3h+5h=15$

$8h=15$

$h=\frac{15}{8}$

1st quadrant $x>0,y>0$ and intersection of line $y=x\, and\: y=-x$

will also give point

for $y=x , \; \; \; \; \; \; 3x+5x=15$

for $y=-x , \; \; \; \; \; \; x=\frac{15}{8}$

$3x-5x=15=x=-\frac{15}{2} 2^{nd}\; quadrant$

$x<0, y>0$

as $y=x,y=-x$ all point on these lines are equidistant from coordinate axes so we find there intersection and determine quadrant.

$\\x>0,y>0=I^{st}\\\; \; \; \\ x<0,y>0=2^{nd}$

Option 1)

$1^{st}$ quadrant

Option 2)

$1^{st},2^{nd}and \; 4^{th}$ quadrants

Option 3)

$4^{th}$ quadrant

Option 4)

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

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