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If each of the points $(a, 4),(-2, b)$ lies on the line joining the points $(2,-1)$ and $(5,-3)$ then the point $(a, b)$ lies on the line
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Verified Answer
The correct answer is:
$2 x+6 y+1=0$
Equation of line joining $(2,-1)$ and $(5,-3)$
$\begin{aligned}
& y+1=\frac{-3+1}{5-2}(x-2) \\
\Rightarrow \quad & 3(y+1)=-2(x-2) \\
\Rightarrow \quad & 3 y+3=-2 x+4 \\
L: & 2 x+3 y=1 \\
\because \quad & (a, 4) \text { lies on } L, \\
& 2 a+12=1 \Rightarrow 2 a=-11 \\
\therefore \quad & a=\frac{-11}{2}
\end{aligned}$
$(-2, b)$ lies on $L$ :
$-4+3 b=1 \Rightarrow 3 b=5 \Rightarrow b=\frac{5}{3}$
$\therefore \quad(a, b)$ i.e. $\left(\frac{-11}{2}, \frac{5}{3}\right)$ lies on line $2 x+6 y+1=0$.
$\begin{aligned}
& y+1=\frac{-3+1}{5-2}(x-2) \\
\Rightarrow \quad & 3(y+1)=-2(x-2) \\
\Rightarrow \quad & 3 y+3=-2 x+4 \\
L: & 2 x+3 y=1 \\
\because \quad & (a, 4) \text { lies on } L, \\
& 2 a+12=1 \Rightarrow 2 a=-11 \\
\therefore \quad & a=\frac{-11}{2}
\end{aligned}$
$(-2, b)$ lies on $L$ :
$-4+3 b=1 \Rightarrow 3 b=5 \Rightarrow b=\frac{5}{3}$
$\therefore \quad(a, b)$ i.e. $\left(\frac{-11}{2}, \frac{5}{3}\right)$ lies on line $2 x+6 y+1=0$.
$\begin{aligned}
& y+1=\frac{-3+1}{5-2}(x-2) \\
\Rightarrow \quad & 3(y+1)=-2(x-2) \\
\Rightarrow \quad & 3 y+3=-2 x+4 \\
L: & 2 x+3 y=1 \\
\because \quad & (a, 4) \text { lies on } L, \\
& 2 a+12=1 \Rightarrow 2 a=-11 \\
\therefore \quad & a=\frac{-11}{2}
\end{aligned}$
$(-2, b)$ lies on $L$ :
$-4+3 b=1 \Rightarrow 3 b=5 \Rightarrow b=\frac{5}{3}$
$\therefore \quad(a, b)$ i.e. $\left(\frac{-11}{2}, \frac{5}{3}\right)$ lies on line $2 x+6 y+1=0$.
$\begin{aligned}
& y+1=\frac{-3+1}{5-2}(x-2) \\
\Rightarrow \quad & 3(y+1)=-2(x-2) \\
\Rightarrow \quad & 3 y+3=-2 x+4 \\
L: & 2 x+3 y=1 \\
\because \quad & (a, 4) \text { lies on } L, \\
& 2 a+12=1 \Rightarrow 2 a=-11 \\
\therefore \quad & a=\frac{-11}{2}
\end{aligned}$
$(-2, b)$ lies on $L$ :
$-4+3 b=1 \Rightarrow 3 b=5 \Rightarrow b=\frac{5}{3}$
$\therefore \quad(a, b)$ i.e. $\left(\frac{-11}{2}, \frac{5}{3}\right)$ lies on line $2 x+6 y+1=0$.
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