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Two straight lines u=0 and v=0 passes through the origin forming an angle of \tan ^{-1}(7 / 9) with each other. If the ratio of the slopes of v=0 and u=0 is 9 / 2, then their equations are

Option: 1

y=3 x \text { and } 3 y=2 x


Option: 2

2 y=3 x \text { and } 3 y=x


Option: 3

y+3 x=0 \text { and } 3 y+2 x=0


Option: 4

2 y+3 x=0 \text { and } 3 y+x=0


Answers (1)

best_answer

Let the slope of u=0 be m, then the slope of v=0 is \frac{9 m}{2}.
\begin{array}{ll} \therefore & \frac{7}{9}=\left|\frac{m-\frac{9 m}{2}}{1+m \cdot \frac{9}{m}}\right|=\left|\frac{-7 m}{2+9 m^2}\right| \\ \text { i.e., } & \frac{-7 m}{2+9 m^2}= \pm \frac{7}{9} \\ \text { i.e., } & 2+9 m^2= \pm 9 m \\ \text { i.e., } & 9 m^2-9 m+2=0 \\ \text { or } & 9 m^2+9 m+2=0 \\ & m=\frac{9 \pm \sqrt{81-72}}{18} \\ & =\frac{9 \pm 3}{18}=\frac{2}{3}, \frac{1}{3} \\ & m=\frac{-9 \pm 3}{18}=-\frac{2}{3},-\frac{1}{3} \end{array}

\therefore \text{ Equation of the lines are } \\ (i) \quad3 y=x \quad\text{ and }\quad 2 y=3 x\\ (ii)\quad 3 y=2 x \quad \text{ and }\quad y=3 x\\ (iii)\quad x+3 y=0 \text{ and }\quad 3 x+2 y=0\\ (iv)\quad2 x+3 y=0 \text{ and } \quad 3 x+y=0\\
 

 

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