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What is equation of straight line pass through the point of
intersection of the line $\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{3}+\frac{y}{2}=1$, and parallel the $4 \mathrm{x}+5 \mathrm{y}-6=0 ?$
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intersection of the line $\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{3}+\frac{y}{2}=1$, and parallel the $4 \mathrm{x}+5 \mathrm{y}-6=0 ?$
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Verified Answer
The correct answer is:
$20 \mathrm{x}+25 \mathrm{y}-54=0$
Given equatiosns: $\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{3}+\frac{y}{2}=1$
Point of intersection $=\left(\frac{6}{5}, \frac{6}{5}\right)$
Let equation of line be $4 x+5 y+k=0$
Putting $\left(\frac{6}{5}, \frac{6}{5}\right), k=-\frac{54}{5}$
Equation of line is $20 x+25 y-54=0$
Point of intersection $=\left(\frac{6}{5}, \frac{6}{5}\right)$
Let equation of line be $4 x+5 y+k=0$
Putting $\left(\frac{6}{5}, \frac{6}{5}\right), k=-\frac{54}{5}$
Equation of line is $20 x+25 y-54=0$
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