Suppose $P$ and $Q$ lie on $3 x+4 y-4=0$ and $5 x-y-4=0$ respectively. If the mid-point of $P Q$ is $(1,5)$, then the slope of the line passing through $P$ and $Q$ is

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Aicharya · Accepted Answer

Let the line $P Q$ be. $y-5=m(x-1)$ ...(i) Substituting $y=m x+5-m$ in the equation $5 x-y-4=0$ We have, $.5 x-m x-5+m-4=0$ $\Rightarrow \quad(5-m) x+m-9=0$ Therefore, $x=\frac{9-m}{5-m}$ and $y=m\left(\frac{9-m}{5-m}\right)+5-m$ $=\frac{25-m}{5-m}$ Here, $Q=\left(\frac{9-m}{5-m}, \frac{25-m}{5-m}\right)$ Substituting $y=m x+5-m$ in the equation $3 x+4 y-4=0$ We have, $3 x+4(m x+5-m)-4=0$ $(3+4 m) x+16-4 m=0$ Therefore, $x=\frac{4 m-16}{4 m+3}$ and $\quad y=\frac{m(4 m-16)}{4 m+3}+5-m$ $y=\frac{m+15}{4 m+3}$ Hence, $P=\left(\frac{4 m-16}{4 m+3}, \frac{m+15}{4 m+3}\right)$ Since, $m(1,5)$ is the mid-point of $P Q$, we have $I=\frac{1}{2}\left[\frac{9-m}{5-m}+\frac{4 m-16}{4 m+3}\right]$.......(ii) and $5=\frac{1}{2}\left[\frac{25-m}{5-m}+\frac{m+15}{4 m+3}\right]$.....(iii) From Eq. (ii), we get $\begin{aligned} 2(5-m) & (4 m+3) \\ & =(9-m)(4 m+3)+(5-m)(4 m-16)\end{aligned}$ $\begin{aligned} \Rightarrow 2\left(-4 m^2+17 m+15\right)=\left(-4 m^2\right. & +33 m+27) \\ & +\left(-4 m^2+36 m-80\right)\end{aligned}$ $\Rightarrow-8 m^2+34 m+30=-8 m^2+69 m-53$ $\Rightarrow \quad 35 m=83$ $\Rightarrow \quad m=\frac{83}{35}$

Suppose $P$ and $Q$ lie on $3 x+4 y-4=0$ and $5 x-y-4=0$ respectively. If the mid-point of $P Q$ is $(1,5)$,…

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