Question 20.12.2: Find the equation of the line through the intersection of li......
Find the equation of the line through the intersection of lines 3 x+4 y=7 and x-y+2=0 and whose slope is 5 .
Step-by-Step
The given lines are 3 x+4 y-7=0 and x-y+2=0 .
The equation of any line through the point of intersection of the given lines is of the form.
\begin{aligned}& (3 x+4 y-7)+k(x-y+2)=0\qquad \quad …(i)\\ \\\Rightarrow & (3+k) x+(4-k) y+(2 k-7)=0 \\ \\\Rightarrow & (4-k) y=-(3+k) x+(7-2 k) \\ \\\Rightarrow & y=\frac{-(3+k)}{4-k} x+\frac{(7-2 k)}{(4-k)} \\ \\\Rightarrow & y=\frac{(k+3)}{(k-4)} x+\frac{(7-2 k)}{(4-2 k)}.\end{aligned}Slope of this line is \frac{(k+3)}{(k-4)} .
\therefore \frac{k+3}{k-4}=5 \Rightarrow k+3=5 k-20 \Rightarrow 4 k=23 \Rightarrow \frac{23}{4} \text {. }Substituting k=\frac{23}{4} in (i), we get
\begin{aligned}& (3 x+4 y-7)+\frac{23}{4}(x-y+2)=0\\ \\\Rightarrow & 4(3 x+4 y-7)+23(x-y+2)=0\\ \\\Rightarrow & 35 x-7 y+18=0, \text { which is the required equation. }\end{aligned}
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