Find the equation of the line passing through the point of intersection of the lines `4x+7y-3=0 and 2x-3y+1=0`, which has equal intercepts on the axes.

Find the equation of the line passing through the point of intersectio

1.	Find the equation of the line passing through the point of intersection of the lines `4x+7y-3=0 and 2x-3y+1=0`, which has equal intercepts on the axes.
Answer» Let the point equation of the required line be `(x)/(a)+(y)/(a)=1` i.e. x+y=a …..(i) The equation of the given line are `4x+7y=3......(i)` `2x-3y=-1......(ii)` On solving (ii) and (iii), we get `x=(1)/(13), y=(5)/(13)` So, the point of intersection of the given line is `P((1)/(13),(5)/(13))` Putting `x=(1)/(13) and y=(5)/(13)` in (i), we get `a=((1)/(13)+(5)/(13))=(6)/(13)` Hence, the required equation is `x+y=(6)/(13)` i.e. 13x+13y-6=0

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Find the equation of the line passing through the point of intersection of the lines `4x+7y-3=0 and 2x-3y+1=0`, which has equal intercepts on the axes.

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