Question

Find the Equation of line through the intersection of lines 3x+4y=7 and x-y+2=0 and whose slope is 5

Answer 1

We have to find the equation of a line whose slope is,

$m=5$

First let's find the point of intersection of lines,

$3x+4y=7\quad\longrightarrow\quad(1)$

and,

$x-y+2=0\\\\y=x+2$

Then (1) becomes,

$3x+4(x+2)=7\\\\7x+8=7\\\\x=-\dfrac{1}{7}$

Therefore,

$y=-\dfrac{1}{7}+2\\\\\\y=\dfrac{13}{7}$

So the two lines intersect at the point $\left(-\dfrac{1}{7},\ \dfrac{13}{7}\right).$ Given that our line also passes through this line.

Hence by point - slope formula,

$y-\dfrac{13}{7}=5\left(x+\dfrac{1}{7}\right)\\\\\\7y-13=5(7x+1)\\\\7y-13=35x+5\\\\\mathbf{35x-7y+18=0}$

This is the equation of our line.