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If the line passing through the points $(5,1, a)$ and $(3, b, 1)$ crosses the YZ plane at the point $\left(0, \frac{17}{2}, \frac{-18}{2}\right)$, then $a+b=$
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Verified Answer
The correct answer is:
10
Equation of line through $(5,1, a)$ and $(3, b, 1)$ is $\frac{x-5}{(5-3)}=\frac{y-1}{1-b}=\frac{z-a}{a-1}=\lambda$ (say)
Since this line passess through $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$ Hence $\frac{0-5}{5-3}=\lambda \Rightarrow \lambda=-\frac{5}{2}$
$$
\begin{aligned}
& \text { and } \frac{\frac{17}{2}-1}{1-b}=\lambda=\frac{-5}{2} \\
& \Rightarrow b=4 \\
& \text { and } \frac{-\frac{13}{2}-a}{a-1}=\lambda=-\frac{5}{2} \\
& \Rightarrow a=6
\end{aligned}
$$
Therefore $a+b=6+4=10$
Since this line passess through $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$ Hence $\frac{0-5}{5-3}=\lambda \Rightarrow \lambda=-\frac{5}{2}$
$$
\begin{aligned}
& \text { and } \frac{\frac{17}{2}-1}{1-b}=\lambda=\frac{-5}{2} \\
& \Rightarrow b=4 \\
& \text { and } \frac{-\frac{13}{2}-a}{a-1}=\lambda=-\frac{5}{2} \\
& \Rightarrow a=6
\end{aligned}
$$
Therefore $a+b=6+4=10$
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