Question

A line $4x+y=1$ passes through the point $A(2,-7)$ meets the lines $BC$ whose equation is $3x-4y+1=0$ at the point $B$.

The equation of the line $AC$ so that $AB=AC$, is

• A. $52x+89y+519=0$
• B. $52x+89y–519=0$
• C. $89x+52y+519=0$
• D. $89x+52y–519=0$

Answer 1

The correct option is A

$52x+89y+519=0$

Step 1: Draw the diagram according to the given information:

Let $AB$ makes an angle $\alpha$ with $BC$.

Given $AB=AC$

So $\angle ABC=\angle ACB$

Step 2: Finding slope

Equation of$BC$ is $3x–4y+1=0$

Slope of $BC$, $m_1=\frac{3}{4}$

Equation of $AB$ is $4x+y=1$

Slope of $AB$, $m_2=-4$

Step 3: Finding the value for $\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\alpha }$

We know that

$\begin{array}{rcl}\tan \alpha & =& \mathbf{\left|}\frac{\mathbf{(}{\mathbf{m}}_{\mathbf{1}}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{m}}_{\mathbf{2}}\mathbf{)}}{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{{\mathbf{2}}_{}}\mathbf{)}}\mathbf{\right|}\\ & =& \left|\frac{(¾+4)}{(1–(¾\left)4\right)}\right|\\ & =& \left|\frac{\left({\displaystyle \frac{19}{4}}\right)}{2}\right|\\ & =& \frac{19}{8}\end{array}$

Let the slope of $AC$ be $m$

So

$$\begin{gathered} \Rightarrow \tan \alpha =\left|\frac{(m–3/4)}{\left(1+{\displaystyle \frac{3m}{4}}\right)}\right| \\ \Rightarrow \frac{19}{8}=\pm \frac{(4m–3)}{4+3m} \end{gathered}$$

Step 4: Finding the equation

By solving we get $m=-4$ or $m=-\frac{52}{89}$

We have slope of $AB=-4$

So slope of $AC=-\frac{52}{89}$

The equation of $AC$ passing through $(2,-7)$ and slope $-\frac{52}{89}$ is

$\begin{array}{rcl}& \Rightarrow & y–y_{1_{}}=m(x–x_{1_{}})\\ & \Rightarrow & y+7=(-\frac{52}{89})(x–2)\\ & \Rightarrow & 89y+623+52x–104=0\\ & \Rightarrow & \mathbf{52}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{89}\mathbf{}\mathit{y}\mathbf{}\mathbf{+}\mathbf{}\mathbf{519}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\end{array}$

Hence, The correct option is (A).