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  • A line through  meets the lines <img alt="\mathrm{x+3 y+2=0}" src="https

A line through \mathrm{A(-5,-4)} meets the lines \mathrm{x+3 y+2=0}, and \mathrm{x-y-5=0} at the points B, C and D respectively. If

\mathrm{\left(\frac{15}{A B}\right)^2+\left(\frac{10}{A C}\right)^2=\left(\frac{6}{A D}\right)^2 } find the equation of the line.

Option: 1

\mathrm{2 x-3 y+11=0}


Option: 2

\mathrm{2 x+3 y+22=0}


Option: 3

\mathrm{3 x-2 y+22=0}


Option: 4

none of these


Answers (1)

best_answer

Let the line through A(-5,-4) makes an angle \theta with \mathrm{x}- axis then the distance form its equation is

\mathrm{\frac{x+5}{\cos \theta}=\frac{y+4}{\sin \theta }\quad \quad \quad \quad \dots(1)}

If \mathrm{A B=r_l, A C=r_2, A D=r_3}, then for B

\mathrm{\frac{x+5}{\cos \theta}=\frac{y+4}{\sin \theta}=r_i \quad \Rightarrow \quad B \equiv\left(r_i \cos \theta-5, r_i \sin \theta-4\right)}

For C\mathrm{ \frac{x+5}{\cos \theta}=\frac{y+4}{\sin \theta}=r_2 \Rightarrow C \equiv\left(r_2 \cos \theta-5, r_2 \sin \theta-4\right) }

And for D, \mathrm{ \frac{x+5}{\cos \theta}=\frac{y+4}{\cos \theta}=r_3 \Rightarrow D \equiv\left(r_3 \cos \theta-5, r_3 \sin \theta-4\right)}

Given that B, C, D lie on lines \mathrm{ x+3 y+2=0,2 x+y+4=0} and  \mathrm{x-y-5=0} respectively so that

\begin{array}{llll} \mathrm{\left(r_1 \cos \theta-5\right)+3\left(r_1 \sin \theta-4\right)+2=0 }& \Rightarrow & \mathrm{\frac{15}{r_1}=\cos \theta+3 \sin \theta} \quad \dots(2)\\ \mathrm{2\left(r_2 \cos \theta-5\right)+r_2(\sin \theta-4)+4=0 }& \Rightarrow & {\frac{10}{r_2}=2 \cos \theta+\sin \theta} \quad \dots(3)\\ \mathrm{\left(r_2 \cos \theta-5\right)-\left(r_3 \sin \theta-4\right)-5=0 }& \Rightarrow &{ \frac{6}{r_3}=\cos \theta-\sin \theta}\quad \dots(4) \end{array}

and \mathrm{ \left(r_2 \cos \theta-5\right)-\left(r_3 \sin \theta-4\right)-5=0 \Rightarrow \frac{6}{r_3}=\cos \theta-\sin \theta }
From (2), (3) and (4)    \mathrm{ \left(\frac{15}{r_1}\right)^2+\left(\frac{10}{r_2}\right)^2=\left(\frac{6}{r_3}\right)^2 }
\begin{aligned} &\mathrm{ \Rightarrow \quad(\cos \theta+3 \sin \theta)^2+(2 \cos \theta+\sin \theta)^2=(\cos \theta-\sin \theta)^2 }\\ &\mathrm{ \text { or } \quad 4 \cos ^2 \theta+9 \sin ^2 \theta+12 \sin \theta \cos \theta=0 }\\ &\mathrm{ \Rightarrow \quad(2 \cos \theta+3 \sin \theta)^2=0 \Rightarrow \quad \frac{\cos \theta}{-3}=\frac{\sin \theta}{2}=k \quad \text { (say) }}\\ \end{aligned}
Putting \mathrm{\cos \theta=-3 k, \sin \theta=2 k} in (1), the required equation is \mathrm{2 x+3 y+22=0}

Posted by

Ajit Kumar Dubey

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