Question

Joint equation of pair of lines through$(3,-2)$ and parallel to $x^2-4xy+3y^2=0$ is

• A. $x^2+3y^2-4xy-14x+24y+45=0$
• B. $x^2+3y^2+4xy-14x+24y+45=0$
• C. $x^2+3y^2+4xy-14x+24y-45=0$
• D. $x^2+3y^2+4xy-14x-24y-45=0$

Answer 1

The correct option is A

$x^2+3y^2-4xy-14x+24y+45=0$

Explanation for the correct option:

Step 1: Finding the values of $m_1$ and $m_2$

Given the equation is,$x^2-4xy+3y^2=0$

The given line is a combination of two lines, hence it can be written as,

$x^2-4xy+3y^2=0$

$\begin{array}{l}{x}^2-3xy-xy-3y^2=0\\ \Rightarrow x(x-3y)-y(x-3y)=0\\ \Rightarrow (x-3y)(x-y)=0\\ \Rightarrow x-3y=0......\left(i\right)\\ x-y=0......\left(ii\right)\end{array}$

Let the slope of the equation $\left(i\right)$ is $m_1$.

$m_1=-\frac{1}{-3}=\frac{1}{3}\left[\because m=\frac{-\mathrm{coeff}.\mathrm{of}x}{\mathrm{coeff}.\mathrm{of}y}\right]$

Similarly, the slope of the equation $\left(ii\right)$ is $m_2$

$m_2=-\frac{1}{-1}=1\mathbf{}\left[\because m=\frac{-\mathrm{coeff}\mathrm{of}x}{\mathrm{coeff}.\mathrm{of}y}\right]$

Let the line parallel to the equation $\left(i\right)$ be

$x-3y+{\lambda }_1=0.....\left(iii\right)$

Let the line parallel to the equation $\left(ii\right)$ be

$x-y+{\lambda }_2=0.....\left(iv\right)$

These lines pass through a point $\left(3,-2\right)$.

Step 2: Finding the values of ${\lambda }_1$ and ${\lambda }_2$

Substituting the value in equation $\left(iii\right)$ and $\left(iv\right)$ we get,

From equation $\left(iii\right)$

$$\begin{gathered} 3-3\left(-2\right)+{\lambda }_1=0 \\ 3+6+{\lambda }_1=0 \\ {\lambda }_1=-9 \end{gathered}$$

From equation $\left(iv\right)$

$$\begin{gathered} 3-\left(-2\right)+{\lambda }_2=0 \\ 5+{\lambda }_2=0 \\ {\lambda }_2=-5 \end{gathered}$$

Therefore the equation $\left(iii\right)$ will become

$x-3y-9=0$

Similarly, the equation $\left(iv\right)$ will become

$x-y-5=0$

Step 3: Finding the equation of line

Now combined equation of $\left(iii\right)$ and $\left(iv\right)$ will be

$\begin{array}{rcl}\left(x-3y-9\right)\left(x-y-5\right)& =& 0\\ & \Rightarrow & x^2-xy-5x-3xy+3y^2+15y-9x+9y+45=0\\ & \Rightarrow & x^2+3y^2-4xy-14x+24y+45=0\end{array}$

This is the joint equation of pair of lines passing through $\left(3,-2\right)$ and parallel to $x^2-4xy+3y^2=0$.

Hence, option (A) is the correct answer.