MHB Equation of Line Passing Through \(A(-5,-4)\) & \(B, C\) & \(D\)

[sbhatnagar]

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

\[\left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 = \left(\frac{6}{AD}\right) ^2 \]

, find the equation of the line.

 

[Sudharaka]

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

\[\left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 = \left(\frac{6}{AD}\right) ^2 \]

, find the equation of the line.

Hi sbhatnagar, :)

I shall outline the method to solve this problem.

Let \(m\) be the gradient of the line through \(A(-5,-4),\,B(x_{1},y_{1}),\,C(x_{2},y_{2})\mbox{ and }D(x_{3},y_{3})\). Then,

\[\frac{y_i+4}{x_i+5}=m\mbox{ for each }i=1,\,2,\,3\]

Also since the points B, C and D are on the lines given by \(x+3y+2=0\), \(2x+y+4=0\) and \(x-y-5=0\) respectively, we have,

\[x_1+3y_1+2=0\]

\[2x_2+y_2+4=0\]

\[x_{3}-y_{3}-5=0\]

Using the above six equations we can find \(x_1,\,x_{2},\,x_{3},\,y_1,\,y_{2},\mbox{ and }y_{3}\) in terms of \(m\).

Finally using the given equation, \(\left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 = \left(\frac{6}{AD}\right) ^2 \) the value of \(m\) can be found. Hence the equation of the line can be determined.

Kind Regards,
Sudharaka.

 

[sbhatnagar]

My solution can be found in the spoiler.

Spoiler

Let equation of line \(AC\) be

\[\frac{y+4}{\sin \theta}=\frac{x+5}{\cos \theta}=r\]

Let line \(AE\) make angle \(\theta\) with the \(x\)-axis and intersects \(x+3y+2=0\) at \(B\) at a distance \(r_1\) and line \(2x+y+4=0\) at \(C\) at a distance \(r_2\) and line \(x-y-5=0\) at \(D\) at a distance \(r_3\).
\[\therefore \ AB=r_1 , \ AC=r_2, \ AD=r_3\]

View attachment 212​

Putting \(x=r_1\cos \theta -5\) and \(y=r_1\sin \theta -4\) in \(x+3y+2 =0\) we get

\[\begin{aligned}
x+3y+2 &=0 \\
\Rightarrow r_1 \cos \theta -4 +3(r_1 \sin \theta -4)+2 &=0 \\
\Rightarrow r_1 &= \frac{-5-3(4)+2}{\cos \theta +3 \sin \theta} \\
\Rightarrow r_1 &= \frac{15}{\cos \theta +3 \sin \theta} \quad \cdots \text{(i)}

\end{aligned}\]

Similarly,

\[\begin{aligned}
r_2 &= \frac{10}{2\cos \theta + \sin \theta} \quad \cdots \text{(ii})\\
r_3 &= \frac{6}{\cos \theta - \sin \theta} \quad \cdots \text{(iii)}

\end{aligned}\]

But it is given that

\[\begin{aligned} \left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 &= \left(\frac{6}{AD}\right)^2 \\ \Rightarrow \left(\frac{15}{r_1}\right)^2 + \left(\frac{10}{r_2}\right)^2 &= \left(\frac{6}{r_3}\right)^2 \\ \Rightarrow (\cos \theta + 3\sin \theta )^2+(2\cos \theta + \sin \theta)^2 &=(\cos \theta -\sin \theta)^2 \quad [\text{from equations (i), (ii) and (iii)}] \\ 4\cos^2 \theta + 6 \sin ^2 \theta +12 \sin \theta \cos \theta &=0 \\ (2\cos \theta + 3 \sin \theta )^2 &=0 \\ 2\cos \theta + 3 \sin \theta &=0 \\ \tan \theta &=-\frac{2}{3}\end{aligned}\]

On substituting this in the equation of \(AC\), we get

\[y+4=(x+5)\tan \theta \\ \Rightarrow y+4=-\frac{2}{3}(x+5) \\ \Rightarrow 2x+3y+22=0\]