Question 20.12.3: Find the equation of the line through the intersection of li......
Find the equation of the line through the intersection of lines x+2 y-3=0 and 4 x-y+7=0 and which is parallel to the line 5 x+4 y-20=0 .
Step-by-Step
5 x+4 y-20=0 \Rightarrow y=\frac{-5}{4} x+5 \\ \therefore \quad slope of the given line =\frac{-5}{4}
and slope of the required line =\frac{-5}{4} .
Now, the equation of any line through the intersection of the given lines is of the form
\begin{array}{l}(x+2 y-3)+k(4 x-y+7)=0\qquad \qquad …(i) \\ \\\Rightarrow (1+4 k) x+(2-k) y+(7 k-3)=0 \\ \\\Rightarrow (2-k) y=-(1+4 k) x+(3-7 k)\\ \\\Rightarrow y=\frac{-(1+4 k)}{(2-k)} x+\frac{(3-7 k)}{(2-k)} \\ \\\Rightarrow y=\frac{(1+4 k)}{(k-2)} x+\frac{(3-7 k)}{(2-k)} .\\ \\\text { Slope of this line }=\frac{(1+4 k)}{(k-2)} \\ \\\therefore \quad \frac{(1+4 k)}{(k-2)}=\frac{-5}{4} \Rightarrow 4+16 k=-5 k+10\end{array}\\ \\ \Rightarrow 21 k=6 \Rightarrow k=\frac{6}{21}=\frac{2}{7} .Substituting k=\frac{2}{7} in (i), we get
\begin{aligned}& (x+2 y-3)+\frac{2}{7}(4 x-y+7)=0\\ \\\Rightarrow & (7 x+14 y-21)+(8 x-2 y+14)=0 \\ \\\Rightarrow & 15 x+12 y-7=0, \text { which is the required equation. }\end{aligned}
Related Answered Questions
Question: 20.8.3
Verified Answer:
We have
\begin{aligned}\sqrt{3} x+y+2=0 \Ri...
Question: 20.12.6
Verified Answer:
The given lines are 2 x+3 y-4=0 a...
Question: 20.12.5
Verified Answer:
The equation of any line through the point of inte...
Question: 20.12.4
Verified Answer:
5 x-4 y+1=0 \Rightarrow y=\frac{5}{4} x...
Question: 20.12.2
Verified Answer:
The given lines are 3 x+4 y-7=0 a...
Question: 20.12.1
Verified Answer:
The equation of any line through the point of inte...
Question: 20.11.5
Verified Answer:
Let the origin O be shifted to a ...
Question: 20.11.4
Verified Answer:
Let the origin O be shifted to th...
Question: 20.11.3
Verified Answer:
Let O^{\prime}(h, k) be the point...
Question: 20.11.2
Verified Answer:
Let the original coordinates be P(x, y) [/...