KCET · Maths · Straight Lines
The equation of the line parallel to the line \( 3 x-4 y+2=0 \) and passing through
\( (-2,3) \) is
- A \( 3 x-4 y+18=0 \)
- B \( 3 x-4 y-18=0 \)
- C \( 3 x+4 y+18=0 \)
- D \( 3 x+4 y-18=0 \)
Answer & Solution
Correct Answer
(A) \( 3 x-4 y+18=0 \)
Step-by-step Solution
Detailed explanation
Given line, \(3 x-4 y+2=0 \rightarrow(1)\)
\(\Rightarrow y=\frac{3}{4} x+\frac{1}{2}\)
So, \(m=\frac{3}{4}\) and \(c=\frac{1}{2}\)
Since line is parallel then slope of both lines should same. Then, equation of parallel is given by \(\left(y-y_{1}\right)=\frac{3}{4}\left(x-x_{1}\right)\)
And this line passes through point \((-2,3)\). Then,
\((y-3)=\frac{3}{4}(x-(-2))\)
\(\Rightarrow 4 y+12=3 x+6\)
\(\Rightarrow 3 x-4 y+18=0\)
Therefore, required equation of line is \(3 x-4 y+18=0\)
\(\Rightarrow y=\frac{3}{4} x+\frac{1}{2}\)
So, \(m=\frac{3}{4}\) and \(c=\frac{1}{2}\)
Since line is parallel then slope of both lines should same. Then, equation of parallel is given by \(\left(y-y_{1}\right)=\frac{3}{4}\left(x-x_{1}\right)\)
And this line passes through point \((-2,3)\). Then,
\((y-3)=\frac{3}{4}(x-(-2))\)
\(\Rightarrow 4 y+12=3 x+6\)
\(\Rightarrow 3 x-4 y+18=0\)
Therefore, required equation of line is \(3 x-4 y+18=0\)
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