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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 that has equal intercepts on the axes.

Q: 12     Find the equation of the line passing through the point of intersection of the lines  4x+7y-3=0  and  2x-3y+1=0  that has equal intercepts on the axes.
 

Answers (1)

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Point of intersection of the lines  4x+7y-3=0  and  2x-3y+1=0 is  \left ( \frac{1}{13},\frac{5}{13} \right )
We know that the intercept form of the line is 
\frac{x}{a}+\frac{y}{b}= 1
It is given that line make equal intercepts on x and  y axis
Therefore,
a = b
Now, the equation reduces to
x+y = a         -(i)
It passes through point  \left ( \frac{1}{13},\frac{5}{13} \right )  
Therefore,
a = \frac{1}{13}+\frac{5}{13}= \frac{6}{13}
Put the value of a in equation (i)
we will get
13x+13y=6
Therefore, equation of line is   13x+13y=6


 

Posted by

Gautam harsolia

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