Question

Q.70 (i) find the equation to the straight lines passing through the point (2,3) inclined at angle of 45$°$ to the line 3x+y-5 = 0 (ii) Find the equation of the straight line passing through the point (3,2 ) making an angle of 45$°$ with the line x-2y = 3

Answer 1

Dear student
(b)
$$\begin{gathered} \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{any}\mathrm{line}\mathrm{thrugh}\mathrm{P}(3,2)\mathrm{is} \\ \mathrm{y}-2=\mathrm{m}(\mathrm{x}-3)...\left(1\right) \\ \mathrm{where}\mathrm{m}\mathrm{is}\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{line}. \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{the}\mathrm{line}\left(1\right)\mathrm{makes}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{of}45°\mathrm{with}\mathrm{the}\mathrm{line}\mathrm{x}-2\mathrm{y}-3=0 \\ \mathrm{Therefore},\tan 45°=\pm \left(\frac{\mathrm{m}-{\displaystyle \frac{1}{2}}}{1+{\displaystyle \frac{1}{2}}\mathrm{m}}\right)\left[\mathbf{Using}\mathbf{}\mathbf{tan\theta }\mathbf{=}\mathbf{\pm }\left(\frac{{\mathbf{m}}_{\mathbf{1}}\mathbf{-}{\mathbf{m}}_{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}\right)\right] \\ \Rightarrow 1=\pm \left(\frac{2\mathrm{m}-1}{2+\mathrm{m}}\right) \\ \Rightarrow 2+\mathrm{m}=\pm \left(2\mathrm{m}-1\right) \\ \Rightarrow \boxed{\mathbf{m}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}} \\ \mathrm{Putting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{m}\mathrm{in}\left(1\right),\mathrm{the}\mathrm{required}\mathrm{lines}\mathrm{are} \\ \mathrm{y}-2=3(\mathrm{x}-3)\mathrm{and}\mathrm{y}-2=\frac{-1}{3}(\mathrm{x}-3) \\ \mathrm{or}\boxed{\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{7}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{and}\mathbf{}\mathbf{x}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{-}\mathbf{9}\mathbf{=}\mathbf{0}} \end{gathered}$$
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