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Q.
Find the equation of the line through the intersection of $3x + 4y - 7 = 0$ and $x = y - 2$, and slope $5$.
Straight Lines
Solution:
Given lines are $3x + 4y - 7 = 0$ and $x -y + 2 = 0$
Any line through the intersection of these lines is
$3x + 4y - 7 + k \left( x - y + 2\right) = 0\quad\ldots\left(i\right)$
which can be written as
$\left(3 + k\right)x + \left(4 - k\right)y - 7 + 2k = 0$
Slope of this line $=-\left(\frac{3+k}{4-k}\right)$
$\therefore $ We must have $-\left(\frac{3+k}{4-k}\right)=5$
$\Rightarrow -3-k=20-5k$
$\Rightarrow 4k=23$
$\Rightarrow k=\frac{23}{4}$.
Substituting this value of $k$ in $\left(i\right)$, we get
$3x+4y-7+\frac{23}{4}\left(x-y+2\right)=0$
$\Rightarrow 35x - 7y + 18 = 0$