Question

The pair of straight lines pass through the point $(1,2)$ and perpendicular to the pair of straight lines $3{{x}^{2}}-8xy+5{{y}^{2}}=0$, is
A. $(5x+3y+11)(x+y+3)=0$
B. $(5x+3y-11)(x+y-3)=0$
C. $(3x+5y-11)(x+y+3)=0$
D. $(3x-5y+11)(x+y-3)=0$

Answer 1

Hint: In this question, we are to find the pair of lines that passes through a given point and is perpendicular to the given pair of lines. For this, we have a predefined formula. By substituting the values into the formula, we get the combined equation of pair of straight lines. So, by simplifying them into factors, we get the required line equations.



Formula Used:The equation of pair of straight lines is
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of second-degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of second-degree in $x$ and $y$.
The equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two lines. They are:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$
The equation to the pair of lines that are passing through the origin and perpendicular to $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ is $b{{x}^{2}}-2hxy+a{{y}^{2}}=0$.
The equation to the pair of lines that are passing through the point $(\alpha,\beta )$ and perpendicular to $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ is $b{{(x-\alpha )}^{2}}-2h(x-\alpha )(y-\beta )+a{{(y-\beta )}^{2}}=0$.



Complete step by step solution:Given pair of line’s equation is
$3{{x}^{2}}-8xy+5{{y}^{2}}=0\text{ }...(1)$
We have the homogenous equation as
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0\text{ }...(2)$
Comparing (1) and (2), we get
$a=3;h=-4;b=5$
It is given that a pair of lines that passes through the point $(1,2)$ and is perpendicular to the equation at (1) is calculated by using the formula,
$b{{(x-\alpha )}^{2}}-2h(x-\alpha )(y-\beta )+a{{(y-\beta )}^{2}}=0$
Where $\alpha =1;\beta =2$
Then,
$\begin{align}
  & b{{(x-\alpha )}^{2}}-2h(x-\alpha )(y-\beta )+a{{(y-\beta )}^{2}}=0 \\
 & \Rightarrow 5{{(x-1)}^{2}}-2(-4)(x-1)(y-2)+3{{(y-2)}^{2}}=0 \\
 & \Rightarrow 5({{x}^{2}}-2x+1)+8(xy-2x-y+2)+3({{y}^{2}}-4y+4)=0 \\
 & \Rightarrow 5{{x}^{2}}-10x+5+8xy-16x-8y+16+3{{y}^{2}}-12y+12=0 \\
 & \Rightarrow 5{{x}^{2}}+8xy+3{{y}^{2}}-26x-20y+33=0\text{ }...(3) \\
\end{align}$
Thus, the obtained equation (3) represents the pair of perpendicular lines to the given pair of lines.
Now, equation (3), is simplified into factors for the individual lines as
$\begin{align}
  & 5{{x}^{2}}+8xy+3{{y}^{2}}-26x-20y+33=0 \\
 & \Rightarrow 5{{x}^{2}}+3xy+5xy+3{{y}^{2}}-11x-15x-11y-9y+33=0 \\
 & \Rightarrow x(5x+3y-11)+y(5x+3y-11)-3(5x+3y-11)=0 \\
 & \Rightarrow (5x+3y-11)(x+y-3)=0 \\
\end{align}$
Thus, the required lines are
$5x+3y-11=0;x+y-3=0$



Option ‘B’ is correct



Note: Here, substituting in the formula, we got evaluated the perpendicular pair of lines equation (combined). But for finding individual line equations, we need to simplify the obtained equation since it is in the general form.