find the equation of straight line passing through (4,5) point and equally inclined to plane 3x=4y+7 and 5y=12x+6

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$$\begin{gathered} \mathrm{Slope} \mathrm{of} \mathrm{any} \mathrm{line} \mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \mathrm{is} -\frac{\mathrm{Coefficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} \mathrm{y}}=-\frac{\mathrm{a}}{\mathrm{b}} \\ {\mathrm{L}}_1: 3\mathrm{x}-4\mathrm{y}-7=0 \\ {\mathrm{m}}_1=\frac{3}{4} \\ {\mathrm{L}}_2: 12\mathrm{x}-5\mathrm{y}+6=0 \\ {\mathrm{m}}_2=\frac{12}{5} \\ \mathrm{Let} \mathrm{slope} \mathrm{of} \mathrm{required} \mathrm{line} \mathrm{L} \mathrm{be} \mathrm{m}. \\ \mathrm{As} \mathrm{line} \mathrm{L} \mathrm{is} \mathrm{equally} \mathrm{inclined} \mathrm{to} {\mathrm{L}}_{1 }\mathrm{and} {\mathrm{L}}_2 , \mathrm{hence} \mathrm{it} \mathrm{makes} \mathrm{equal} \mathrm{angle} \mathrm{\theta } \mathrm{with} \mathrm{both} \mathrm{the} \mathrm{lines}. \\ \Rightarrow \left|\frac{\mathrm{m}-{\mathrm{m}}_1}{1+{\mathrm{mm}}_1}\right|=\left|\frac{\mathrm{m}-{\mathrm{m}}_2}{1+{\mathrm{mm}}_2}\right| \\ \Rightarrow \left|\frac{\mathrm{m}-{\displaystyle \frac{3}{4}}}{1+\mathrm{m}.{\displaystyle \frac{3}{4}}}\right|=\left|\frac{\mathrm{m}-{\displaystyle \frac{12}{5}}}{1+\mathrm{m}.{\displaystyle \frac{12}{5}}}\right| \\ \Rightarrow \left|\frac{4\mathrm{m}-3}{3\mathrm{m}+4}\right|=\left|\frac{5\mathrm{m}-12}{12\mathrm{m}+5}\right| \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{if} \left|\mathrm{x}\right|=\left|\mathrm{a}\right|, \mathrm{then} \mathrm{x}=\pm \mathrm{a} \\ \Rightarrow \frac{4\mathrm{m}-3}{3\mathrm{m}+4}=\pm \left(\frac{5\mathrm{m}-12}{12\mathrm{m}+5}\right) \\ \mathrm{Case} \left(\mathrm{i}\right) \\ \frac{4\mathrm{m}-3}{3\mathrm{m}+4}=\left(\frac{5\mathrm{m}-12}{12\mathrm{m}+5}\right) \\ \Rightarrow \left(4\mathrm{m}-3\right)\left(12\mathrm{m}+5\right)=\left(5\mathrm{m}-12\right)\left(3\mathrm{m}+4\right) \\ \Rightarrow 48{\mathrm{m}}^2+20\mathrm{m}-36\mathrm{m}-15=15{\mathrm{m}}^2+20\mathrm{m}-36\mathrm{m}-48 \\ \Rightarrow 48{\mathrm{m}}^2+20\mathrm{m}-36\mathrm{m}-15-15{\mathrm{m}}^2-20\mathrm{m}+36\mathrm{m}+48=0 \\ \Rightarrow 33{\mathrm{m}}^2+33=0 \\ \Rightarrow 33{\mathrm{m}}^2=-33 \\ \mathrm{But} {\mathrm{m}}^2 \mathrm{cannot} \mathrm{be} \mathrm{negative}. \mathrm{Hence} \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{real} \mathrm{value} \mathrm{of} \mathrm{m} \mathrm{that} \mathrm{satisfies} \mathrm{above} \mathrm{equation}. \\ \frac{4\mathrm{m}-3}{3\mathrm{m}+4}=-\left(\frac{5\mathrm{m}-12}{12\mathrm{m}+5}\right) \\ \Rightarrow \left(4\mathrm{m}-3\right)\left(12\mathrm{m}+5\right)=-\left(5\mathrm{m}-12\right)\left(3\mathrm{m}+4\right) \\ \Rightarrow 48{\mathrm{m}}^2+20\mathrm{m}-36\mathrm{m}-15=-15{\mathrm{m}}^2-20\mathrm{m}+36\mathrm{m}+48 \\ \Rightarrow 48{\mathrm{m}}^2+20\mathrm{m}-36\mathrm{m}-15+15{\mathrm{m}}^2+20\mathrm{m}-36\mathrm{m}-48=0 \\ \Rightarrow 63{\mathrm{m}}^2-32\mathrm{m}-63=0 \\ \Rightarrow 63{\mathrm{m}}^2-81\mathrm{m}+49\mathrm{m}-63=0 \\ \Rightarrow 9\mathrm{m}\left(7\mathrm{m}-9\right)+7\left(7\mathrm{m}-9\right)=0 \\ \Rightarrow \left(7\mathrm{m}-9\right)\left(9\mathrm{m}+7\right)=0 \\ \Rightarrow \mathrm{Either} \mathrm{m}=\frac{9}{7} \mathrm{and} \mathrm{m}=-\frac{7}{9} \\ \mathrm{Equation} \mathrm{of} \mathrm{line} \mathrm{having} \mathrm{slope} \mathrm{m} \mathrm{and} \mathrm{passing} \mathrm{through} \mathrm{points} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \mathrm{is} \\ \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \mathrm{Line} \mathrm{L} \mathrm{passes} \mathrm{through} \left(4,5\right) \\ \mathrm{When} \mathrm{m}=\frac{9}{7} \\ \mathrm{L}: \mathrm{y}-5=\frac{9}{7}\left(\mathrm{x}-4\right) \\ \Rightarrow 7\mathrm{y}-35=9\mathrm{x}-36 \\ \Rightarrow 9\mathrm{x}-36-7\mathrm{y}+35=0 \\ \mathbf{\Rightarrow }\mathbf{9}\mathbf{x}\mathbf{-}\mathbf{7}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{ }\left(\mathbf{Answer}\right) \\ \mathrm{When} \mathrm{m}=-\frac{7}{9} \\ \mathrm{L}: \mathrm{y}-5=-\frac{7}{9}\left(\mathrm{x}-4\right) \\ \Rightarrow 9\mathrm{y}-55=-7\mathrm{x}+28 \\ \Rightarrow 7\mathrm{x}+9\mathrm{y}-45-28=0 \\ \mathbf{\Rightarrow }\mathbf{7}\mathbf{x}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{-}\mathbf{73}\mathbf{=}\mathbf{0}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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