Question 20.12.4: Find the equation of the line through the intersection of th......
Find the equation of the line through the intersection of the lines 3 x+y-9=0 and 4 x+3 y-7=0 and which is perpendicular to the line 5 x-4 y+1=0 .
Step-by-Step
5 x-4 y+1=0 \Rightarrow y=\frac{5}{4} x+\frac{1}{4}
Slope of the given line is \frac{5}{4} .
Now, the equation of any line through the intersection of the given lines is of the form
\begin{aligned}& (3 x+y-9)+k(4 x+3 y-7)=0 \qquad \quad …(i)\\ \\\Rightarrow & (3+4 k) x+(1+3 k) y-(9+7 k)=0\\ \\\Rightarrow & (1+3 k) y=-(3+4 k) x+(9+7 k)\\ \\\Rightarrow & y=-\frac{(3+4 k)}{(1+3 k)} x+\frac{(9+7 k)}{(1+3 k)}\qquad \qquad …(ii)\end{aligned}Let m be the slope of the line perpendicular to the required line.
Then, m \times \frac{5}{4}=-1 \Rightarrow m=\frac{-4}{5} .
\therefore \quad we must have \frac{-(3+4 k)}{(1+3 k)}=\frac{-4}{5} \\ \Rightarrow 15+20 k=4+12 k \Rightarrow 8 k=-11 \Rightarrow k=\frac{-11}{8} .
Substituting k=\frac{-11}{8} in (i), we get
\begin{aligned}& (3 x+y-9)-\frac{11}{8}(4+3 y-7)=0 \\ \\\Rightarrow & (24 x+8 y-72)-44 x-33 y+77=0 \\ \\\Rightarrow & 20 x+25 y-5=0 \Rightarrow 4 x+5 y-1=0 .\end{aligned}Related Answered Questions
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