KCET · Maths · Application of Derivatives
A line passes through \( (2,2) \) and is perpendicular in the line \( 3 x+y=3 \) its \( y \)-intercepts is
- A \( \frac{1}{3} \)
- B \( \frac{2}{3} \)
- C \( \frac{4}{3} \)
- D \( 11 \)
Answer & Solution
Correct Answer
(C) \( \frac{4}{3} \)
Step-by-step Solution
Detailed explanation
Given line, \(3 x+y=3\)
\(\Rightarrow y=-3 x+3\)
So, slope of this line is \(m_{1}=-3\)
Then, slope of line perpendicular to this line is \(m_{2}=\frac{1}{3}\)
General equation line passing through point \(\left(x_{1}, y_{1}\right)\) is
\(\left(y-y_{1}\right)=m\left(x-x_{1}\right)\)
Here \(m=\frac{1}{3}\) and \(\left(x_{1}, y_{1}\right)=(2,2)\). So,
\(y-2=\frac{1}{3}(x-2)\)
\(\Rightarrow y-2=\frac{x}{3}-\frac{2}{3}\)
\(\Rightarrow y=\frac{x}{3}+2-\frac{2}{3}\)
\(\Rightarrow y=\frac{x}{3}+\frac{4}{3}\)
So, \(y\)-axis intercept is, \(\frac{4}{3}\)
\(\Rightarrow y=-3 x+3\)
So, slope of this line is \(m_{1}=-3\)
Then, slope of line perpendicular to this line is \(m_{2}=\frac{1}{3}\)
General equation line passing through point \(\left(x_{1}, y_{1}\right)\) is
\(\left(y-y_{1}\right)=m\left(x-x_{1}\right)\)
Here \(m=\frac{1}{3}\) and \(\left(x_{1}, y_{1}\right)=(2,2)\). So,
\(y-2=\frac{1}{3}(x-2)\)
\(\Rightarrow y-2=\frac{x}{3}-\frac{2}{3}\)
\(\Rightarrow y=\frac{x}{3}+2-\frac{2}{3}\)
\(\Rightarrow y=\frac{x}{3}+\frac{4}{3}\)
So, \(y\)-axis intercept is, \(\frac{4}{3}\)
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