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

  • 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

Answers (1)

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

So 1st and 2nd quadrant:

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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