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# A line is such that its segment between the straight lines $5x-y+4=0$ and $3x+4y-4=0$ is bisected at the point $\left( 1,5 \right)$ . Obtain its equation.(a) $107x+y-92=0$ (b) $17x-3y+92=0$ (c) $10x+3y+92=0$ (d) $107x-3y-92=0$

Last updated date: 09th Aug 2024
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Hint:
We start solving the problem by assuming the slope of the required line as m and we then find its equation by using the fact that the equation of the line passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ and having slope m is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ . We then find the points of intersection of the obtained line with the lines $5x-y+4=0$ and $3x+4y-4=0$. We then find the midpoint of the obtained intersection points and equate it to $\left( 1,5 \right)$. We then make the necessary calculations to get the required value of m which then gives us the equation of the line.

According to the problem, we are asked to find the equation of the line is such that its segment between the straight lines $5x-y+4=0$ and $3x+4y-4=0$ is bisected at the point $\left( 1,5 \right)$.
Let us assume the slope of the require line be ‘m’. We know that the equation of the line passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ and having slope m is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ .
So, we get the equation of the required line as $y-5=m\left( x-1 \right)$ .
$\Rightarrow y-5=mx-m$ .
$\Rightarrow mx-y-m+5=0$ ---(1).
Let us find the point of intersection of lines $5x-y+4=0$ and $mx-y-m+5=0$ .
So, we get $\dfrac{x}{\left( -1\times \left( -m+5 \right) \right)-\left( -1\times 4 \right)}=\dfrac{y}{\left( 4\times m \right)-\left( \left( -m+5 \right)\times 5 \right)}=\dfrac{1}{\left( 5\times -1 \right)-\left( -1\times m \right)}$ .
$\Rightarrow \dfrac{x}{\left( m-5 \right)-\left( -4 \right)}=\dfrac{y}{\left( 4m \right)-\left( -5m+25 \right)}=\dfrac{1}{\left( -5 \right)-\left( -m \right)}$ .
$\Rightarrow \dfrac{x}{m-5+4}=\dfrac{y}{4m+5m-25}=\dfrac{1}{-5+m}$ .
$\Rightarrow \dfrac{x}{m-1}=\dfrac{y}{9m-25}=\dfrac{1}{m-5}$ .
$\Rightarrow x=\dfrac{m-1}{m-5}$ , $y=\dfrac{9m-25}{m-5}$ .
So, the intersection point of $5x-y+4=0$ and $mx-y-m+5=0$ is $A\left( \dfrac{m-1}{m-5},\dfrac{9m-25}{m-5} \right)$ ---(2).
Now, let us find the point of intersection of lines $3x+4y-4=0$ and $mx-y-m+5=0$ .
So, we get $\dfrac{x}{\left( 4\times \left( -m+5 \right) \right)-\left( -1\times -4 \right)}=\dfrac{y}{\left( -4\times m \right)-\left( \left( -m+5 \right)\times 3 \right)}=\dfrac{1}{\left( 3\times -1 \right)-\left( 4\times m \right)}$ .
$\Rightarrow \dfrac{x}{-4m+20-4}=\dfrac{y}{-4m+3m-15}=\dfrac{1}{-3-4m}$ .
$\Rightarrow \dfrac{x}{-4m+16}=\dfrac{y}{-m-15}=\dfrac{1}{-3-4m}$ .
$\Rightarrow x=\dfrac{4m-16}{3+4m}$ , $y=\dfrac{m+15}{3+4m}$ .
So, the intersection point of $3x+4y-4=0$ and $mx-y-m+5=0$ is $B\left( \dfrac{4m-16}{3+4m},\dfrac{m+15}{3+4m} \right)$ ---(3).
According to the problem, we are given that the mid-point of AB is $\left( 1,5 \right)$ .
So, we have $\left( \dfrac{\left( \dfrac{m-1}{m-5}+\dfrac{4m-16}{3+4m} \right)}{2},\dfrac{\left( \dfrac{9m-25}{m-5}+\dfrac{m+15}{3+4m} \right)}{2} \right)=\left( 1,5 \right)$ .
$\Rightarrow \left( \dfrac{\left( \dfrac{\left( m-1 \right)\left( 3+4m \right)+\left( 4m+16 \right)\left( m-5 \right)}{\left( m-5 \right)\left( 3+4m \right)} \right)}{2},\dfrac{\left( \dfrac{\left( 9m-25 \right)\left( 3+4m \right)+\left( m+15 \right)\left( m-5 \right)}{\left( m-5 \right)\left( 3+4m \right)} \right)}{2} \right)=\left( 1,5 \right)$ .
$\Rightarrow \left( \dfrac{\left( \dfrac{4{{m}^{2}}+3m-4m-3+4{{m}^{2}}-16m-20m+80}{4{{m}^{2}}+3m-20m-15} \right)}{2},\dfrac{\left( \dfrac{36{{m}^{2}}+27m-100m-75+{{m}^{2}}+15m-5m-75}{3m+4{{m}^{2}}-15-20m} \right)}{2} \right)=\left( 1,5 \right)$ .
$\Rightarrow \left( \dfrac{\left( \dfrac{8{{m}^{2}}-37m+77}{4{{m}^{2}}-17m-15} \right)}{2},\dfrac{\left( \dfrac{37{{m}^{2}}-63m-150}{4{{m}^{2}}-17m-15} \right)}{2} \right)=\left( 1,5 \right)$ .
$\Rightarrow \dfrac{\left( \dfrac{8{{m}^{2}}-37m+77}{4{{m}^{2}}-17m-15} \right)}{2}=1$, $\dfrac{\left( \dfrac{37{{m}^{2}}-63m-150}{4{{m}^{2}}-17m-15} \right)}{2}=5$.
$\Rightarrow \dfrac{8{{m}^{2}}-37m+77}{4{{m}^{2}}-17m-15}=2$, $\dfrac{37{{m}^{2}}-63m-150}{4{{m}^{2}}-17m-15}=10$.
$\Rightarrow 8{{m}^{2}}-37m+77=8{{m}^{2}}-34m-30$, $37{{m}^{2}}-63m-150=40{{m}^{2}}-170m-150$.
$\Rightarrow 3m=107$, $3{{m}^{2}}-107m=0$.
$\Rightarrow m=\dfrac{107}{3}$, $m\left( 3m-107 \right)=0$.
$\Rightarrow m=\dfrac{107}{3}$, $m=0\text{ or }m=\dfrac{107}{3}$.
We take the common value of m as it should be valid for both coordinates of the midpoint.
So, the value of m is $\dfrac{107}{3}$ . Let us substitute this in equation (1).
So, we get the equation of the line as $\left( \dfrac{107}{3} \right)x-y-\left( \dfrac{107}{3} \right)+5=0$ .
$\Rightarrow \dfrac{107x-3y-107+15}{3}=0$ .
$\Rightarrow 107x-3y-92=0$ .
$\therefore,$ The correct option for the given problem is (d).

Note:
We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully in order to avoid confusion and calculation mistakes. We can also solve this problem by taking the parametric point on both the given lines and then equating the midpoint of those two parametric points to $\left( 1,5 \right)$ to find the parameters which later helps us to find the equation of a line. Similarly, we can expect problems to find the equation of the line if $\left( 1,2 \right)$ divides the line $1:2$ internally.