MHT CET · Maths · Three Dimensional Geometry
The equation of the lines passing through the point \((3,2)\) and making an acute angle of \(45^{\circ}\) with the lien \(x-2 y-3=0\) are
- A \(x+2 y-7=0,2 x-y-4=0\)
- B \(3 x+y-11=0, x+3 y-9=0\)
- C \(3 x-y-7=0, x+3 y-9=0\)
- D \(3 x+y-11=0, x+3 y+9=0\)
Answer & Solution
Correct Answer
(C) \(3 x-y-7=0, x+3 y-9=0\)
Step-by-step Solution
Detailed explanation
Let the slope of required line by \(\mathrm{m}\) then \(\tan 45^{\circ}=\left|\frac{m-\frac{1}{2}}{1+m \times \frac{1}{2}}\right|\)
\(\begin{aligned} & \Rightarrow \pm 1=\frac{m-\frac{1}{2}}{1+m \times \frac{1}{2}} \\ & \Rightarrow m=3 \text { or } m=-\frac{1}{3}\end{aligned}\)
So, required equation
\(\begin{aligned} & (y-2)=3(x-3) \text { or }(y-2)=\frac{-1}{3}(x-3) \\ & \Rightarrow 3 x-y-7=0 \text { or } x+3 y-9=0\end{aligned}\)
\(\begin{aligned} & \Rightarrow \pm 1=\frac{m-\frac{1}{2}}{1+m \times \frac{1}{2}} \\ & \Rightarrow m=3 \text { or } m=-\frac{1}{3}\end{aligned}\)
So, required equation
\(\begin{aligned} & (y-2)=3(x-3) \text { or }(y-2)=\frac{-1}{3}(x-3) \\ & \Rightarrow 3 x-y-7=0 \text { or } x+3 y-9=0\end{aligned}\)
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