Question

The straight lines $3x+4y-5=0$ and $4x-3y=15$ intersected at the point $P.$

On these lines, the points $Q$ and $R$ are chosen so that $PQ=PR.$

The slopes of the line $QR$ passing through $(1,2)$ are

• A. $\left(-7,\frac{1}{7}\right)$
• B. $\left(7,\frac{1}{7}\right)$
• C. $\left(7,-\frac{1}{7}\right)$
• D. $\left(3,-\frac{1}{3}\right)$

Answer 1

The correct option is A

$\left(-7,\frac{1}{7}\right)$

Explanation for the correct option:

Step 1: Find the nature of the given lines.

Given lines are $3x+4y-5=0..\left(i\right)$ and $4x-3y=15..\left(ii\right)$

We know that the standard equation of the line is $Ax+By+C=0$ and the slope $m$ of the line is given by $m=-\frac{A}{B}$.

The slope of line $\left(i\right)$ is $m_1=-\frac{3}{4}$.

The slope of line $\left(ii\right)$ is $m_2=\frac{4}{3}$.

Here, $m_1{m}_2=-1$.

We know that the product of the slope of perpendicular lines is always $-1$.

So the lines are perpendicular.

Step 2: Compute the slope of the required line.

Given $PQ=PR.$

So triangle $PQR$ is isosceles.

Let the slope of $QR=m$.

We know that the angle $\theta$ between two lines can be given by $\tan \theta =\left|\frac{m_2-m}{1+m_{2}m}\right|$

Then,

$$\begin{gathered} \tan 45°=\left|\frac{{\displaystyle \frac{4}{3}}-m}{1+\left({\displaystyle \frac{4}{3}}\right)m}\right| \\ \Rightarrow 1=\pm \frac{4-3m}{3+4m} \end{gathered}$$

When, $1=\frac{4-3m}{3+4m}$

$$\begin{gathered} \Rightarrow 3+4m=4-3m \\ \Rightarrow 7m=1 \end{gathered}$$

So $m=\frac{1}{7}$

When $1=-\frac{4-3m}{3+4m}$

$$\begin{gathered} 3+4m+4-3m=0 \\ m=-7 \end{gathered}$$

Therefore, the value of the slopes of $QR$ is $(-7,\frac{1}{7})$.

Hence option (A) is the correct answer.