Question

A line passes through the point of intersection of the lines $100x+50y-1=0$and $75x+25y+3=0$ and makes an equal intercept on the axes. Its equation is

• A. $25x+25y–1=0$
• B. $5x–5y+3=0$
• C. $25x+25y–4=0$
• D. $25x–25y+6=0$

Answer 1

The correct option is C

$25x+25y–4=0$

Explanation for the correct option:

Step 1. Draw the figure according to the question:

Given the equation of lines are

$$\begin{gathered} 100x+50y-1=0...\left(i\right) \\ 75x+25y+3=0...\left(ii\right) \end{gathered}$$

Step 2. Solving (i) and (ii) to get the point of intersection of the lines.

$\begin{array}{rcl}(150x+50y+6)-\left(100x+50y-1\right)& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}50x+7& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& \frac{-7}{50}\end{array}$

and $\begin{array}{rcl}y& =& \frac{3}{10}\end{array}$

Step 3. Finding the equation of the line

Let the equation of the line which passes through $\begin{array}{rcl}& & \left(\frac{-7}{50},\frac{3}{10}\right)\end{array}$ and having equal intercepts is

$\begin{array}{rcl}\frac{x}{a}+\frac{y}{a}& =& 1\end{array}$

$\Rightarrow$$\begin{array}{rcl}x+y& =& a\end{array}$

$\because$The value $\begin{array}{rcl}& & \left(\frac{-7}{50},\frac{3}{10}\right)\end{array}$ satisfy the equation.

So,

$\begin{array}{c}\frac{-7}{50}+\frac{3}{10}=a\end{array}$

$\Rightarrow$$\begin{array}{c}-7+15=50a\end{array}$

$\Rightarrow$ $8=50a$

$\Rightarrow$ $\frac{4}{25}=a$

$\therefore$the required equation is $x+y=\frac{4}{25}$

$\Rightarrow$ $\mathbf{25}\mathit{x}\mathbf{+}\mathbf{25}\mathit{y}\mathbf{-}\mathbf{4}\mathbf{=}\mathbf{0}$

Hence option C is the answer.