Question

A line passes through the point of intersection of the lines $3x+y+1=0$ and $2x-y+3=0$ and makes equal intercepts with axes. Then, equation of the line is

• A. $5x+5y–3=0$
• B. $x+5y–3=0$
• C. $5x–y–3=0$
• D. $5x+5y+3=0$

Answer 1

The correct option is A

$5x+5y–3=0$

The explanation for the correct option:

Step 1: Expressing the given information.

Given

$\begin{array}{rcl}3x+y+1& =& 0\end{array}$…..(i)

$\begin{array}{rcl}2x–y+3& =& 0\end{array}$ ..…(ii)

Step 2: Finding slope

By solving (i) and (ii) we get the point of intersection of the lines.

So $x=-\frac{4}{5}$ and $y=\frac{7}{5}$

The point of intersection is $\left(-\frac{4}{5},\frac{7}{5}\right)$

Given that the line makes equal intercepts with axes.

So slope could be

$\tan \theta =1or-1$

Step 3: Finding the equation for different slopes

Case 1: Let the slope $=1$

Then the equation of the line is

$\begin{array}{rcl}y–\frac{7}{5}& =& x+\frac{4}{5}\mathbf{}\mathbf{}\mathit{y}\mathbf{-}{\mathit{y}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\left(x-x_1\right)\\ 5x–5y+11& =& 0\end{array}$

Case 2: Let the slope $=-1$

Then the equation of the line is

$\begin{array}{rcl}y–\frac{7}{5}& =& -(x+\frac{4}{5})\mathbf{}\mathbf{}\mathit{y}\mathbf{}\mathbf{–}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{m}\mathbf{(}\mathbf{x}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{x}}_{\mathbf{1}}\mathbf{)}\mathbf{}\\ 5y–7+5x+4& =& 0\\ \mathbf{}\mathbf{5}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{5}\mathit{y}\mathbf{}\mathbf{–}\mathbf{}\mathbf{3}\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{0}\end{array}$

Hence, option (A) is the correct answer.