KCET · Maths · Vector Algebra
Equation of line passing through the point \( (1,2) \) and perpendicular to the line
\( y=3 x-1 \)
- A \( x+3 y-7=0 \)
- B \( x+3 y+7=0 \)
- C \( x+3 y=0 \)
- D \( x-3 y=0 \)
Answer & Solution
Correct Answer
(A) \( x+3 y-7=0 \)
Step-by-step Solution
Detailed explanation
Given line, \(y=3 x-1 \rightarrow(1)\)
General equation of line is given by
\(y=m x+c\)
Here \(m=3\) and \(c=-1\).
Condition for two perpendicular lines is
\(m_{1} \cdot m_{2}=-1\)
So, \(3 m_{2}=-1 \Rightarrow m_{2}=-\frac{1}{3}\)
Standard equation of line passing through point \(\left(x_{1}, y_{1}\right)\) is given by \(\left(y-y_{1}\right)=m\left(x-x_{1}\right)\)
Here \(\left(x_{1}, y_{1}\right)=(1,2)\) So,
\((y-2)=-\frac{1}{3}(x-1)\)
\(\Rightarrow-3 y+6=x-1\)
\(\Rightarrow x+3 y-7=0\)
General equation of line is given by
\(y=m x+c\)
Here \(m=3\) and \(c=-1\).
Condition for two perpendicular lines is
\(m_{1} \cdot m_{2}=-1\)
So, \(3 m_{2}=-1 \Rightarrow m_{2}=-\frac{1}{3}\)
Standard equation of line passing through point \(\left(x_{1}, y_{1}\right)\) is given by \(\left(y-y_{1}\right)=m\left(x-x_{1}\right)\)
Here \(\left(x_{1}, y_{1}\right)=(1,2)\) So,
\((y-2)=-\frac{1}{3}(x-1)\)
\(\Rightarrow-3 y+6=x-1\)
\(\Rightarrow x+3 y-7=0\)
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