Question

The pair of straight lines passing through the point $\left(1,2\right)$ and perpendicular to the pair of straight lines $3x^2-8xy+5y^2=0$, is

• A. $\left(5x+3y+11\right)\left(x+y+3\right)$
• B. $\left(5x+3y-11\right)\left(x+y-3\right)$
• C. $\left(5x+3y-11\right)\left(x+y+3\right)$
• D. $\left(5x+3y+11\right)\left(x+y-3\right)$

Answer 1

The correct option is B

$\left(5x+3y-11\right)\left(x+y-3\right)$

Step 1: Find the slope of the given straight line

Given equation of pair of straight line is,

$$\begin{gathered} 3x^2-8xy+5y^2=0 \\ \Rightarrow 3x^2-3xy-5xy+5y^2=0 \\ \Rightarrow 3x\left(x-y\right)-5y\left(x-y\right)=0 \\ \Rightarrow \left(x-y\right)\left(3x-5y\right)=0 \end{gathered}$$

So the equation of the line are

$$\begin{gathered} x-y=0.................\left(i\right) \\ 3x-5y=0...............\left(ii\right) \end{gathered}$$

Slope of $\left(i\right)$ is $1$ and slope of $\left(ii\right)$ is $\frac{3}{5}$

So the lines perpendicular to $\left(i\right)$ and $\left(ii\right)$ will have the slope $-1$ and $-\frac{5}{3}$

Step 2: Find the equation of the lines

Also given these perpendicular lines passes through the point $\left(1,2\right)$

So equation of line passing through $\left(1,2\right)$ and slope $-1$ is

$$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-2=-1\left(x-1\right) \\ \Rightarrow x+y-3=0 \end{gathered}$$

The equation of line passing through $\left(1,2\right)$ and slope $-\frac{5}{3}$ is

$$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-2=-\frac{5}{3}\left(x-1\right) \\ \Rightarrow 5x+3y-11=0 \end{gathered}$$

So the combined equation is $\left(5x+3y-11\right)\left(x+y-3\right)$

Hence, option $\left(\mathrm{B}\right)$ is the correct answer.