# Straight Lines

#### sbhatnagar

##### Active member
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

##### Well-known member
MHB Math Helper
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.

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#### sbhatnagar

##### Active member
My solution can be found in the 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 $$r_3$$.
$\therefore \ AB=r_1 , \ AC=r_2, \ AD=r_3$

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

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