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  • Consider two lines <img alt="\mathrm{L_1: a_1 x+b_1 y+c_1=0 \text { and } L_2: a_2 x+b_2 y+c_2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BL_1%3A%20a_1%20x&plus;b_

Consider two lines \mathrm{L_1: a_1 x+b_1 y+c_1=0 \text { and } L_2: a_2 x+b_2 y+c_2=0} where \mathrm{c_1, c_2 \neq 0} intersecting at a point P. A line \mathrm{L_{3}} is drawn through the origin meeting the line \mathrm{\mathrm{L}_1 \text { and } \mathrm{L}_2} in A and B respectively such that PA = PB. Similarly one more line \mathrm{L}_4 is drawn through the origin meeting the lines \mathrm{\mathrm{L}_1 \text { and } \mathrm{L}_2} in \mathrm{\mathrm{A}_1 \text { and } \mathrm{B}_1} respectively such that \mathrm{\mathrm{PA}_1 = \mathrm{PB}_1}. The combined equation of lines \mathrm{\mathrm{L}_3 \; and \; \mathrm{L}_4} is \mathrm{\left(k_1 x+k_2 y\right)^2\left(a_1^2+b_1^2\right)}, then find \mathrm{\mathrm{k}_1 \text { and } \mathrm{k}_2}

Option: 1

\mathrm{K_1=a_2, K_2=b_2}


Option: 2

\mathrm{K_1=b_2, K_2=a_2}


Option: 3

\mathrm{K_1=a_1, K_2=b_1}


Option: 4

\mathrm{K_1=a_2, K_2=b_1}


Answers (1)

best_answer

Draw lines \mathrm{L_{5}\; and\; L_{6}} through P parallel to \mathrm{L_{3}\; and\; L_{4}} respectively. 

\begin{aligned} & \angle \mathrm{A}_1 \mathrm{PM}=\angle \mathrm{PAB}=\angle \mathrm{PBA} \quad( \mathrm{AP}=\mathrm{BP}) \\ \\& \mathrm{=\angle B_1 P M \text { i.e. } L_5} \\ & \end{aligned}

\Rightarrow PM is the bisector of \mathrm{L_{1}\; and\; L_{2}}

Again \mathrm{\angle \mathrm{APN}=\angle \mathrm{PA}_1 \mathrm{~B}_1=\angle \mathrm{PB}_1 \mathrm{~A}_1=\angle \mathrm{BPN} \quad\left(\mathrm{Q} \mathrm{A}_1 \mathrm{P}=\mathrm{B}_1 \mathrm{P}\right)}

\Rightarrow PN i.e. \mathrm{L_{6}} is the bisector of \mathrm{L_{1}\; and\; L_{2}}

Now joint equation of \mathrm{L_{5}\; and\; L_{6}} is \mathrm{\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}= \pm \frac{a_2 x+b_2 y+c_2}{\sqrt{a_2^2+b_2^2}}}

\Rightarrow  Joint equation of \mathrm{L_{3}\; and\; L_{4}} is  \mathrm{\frac{a_1 x+b_1 y}{\sqrt{a_1^2+b_1^2}}= \pm \frac{a_2 x+b_2 y}{\sqrt{a_2^2+b_2^2}}}    [ \mathrm{L_{3}\; and\; L_{4}} are parallel to \mathrm{L_{5}\; and\; L_{6}} respectively and pass through origin ]

\mathrm{\Rightarrow\left(a_1 x+b_1 y\right)^2\left(a_2^2+b_2^2\right)=\left(a_2 x+b_2 y\right)^2\left(a_1^2+b_1^2\right)}

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Shailly goel

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