The line passing through the points 5,1,a,3,b,1 crosses the  yz-plane at the point  0,172,−132then

# The line passing through the points $\left(5,1,a\right),\left(3,b,1\right)$ crosses the  $yz-$plane at the point  $\left(0,\frac{17}{2},\frac{-13}{2}\right)$then

1. A

$a=2,b=8$

2. B

$a=4,b=6$

3. C

$a=6,b=4$

4. D

$a=8,b=2$

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### Solution:

The $yz-$  plane cuts the line segment joining the points  $A\left({x}_{1},{y}_{1},{z}_{1}\right),B\left({x}_{2},{y}_{2},{z}_{2}\right)$in the ratio$-{x}_{1}:{x}_{2}$

Hence, the   $yz-$plane divides the line segment joining the points $\left(5,1,a\right),\left(3,b,1\right)$in the ratio  $5:3$externally

Hence the point of division is  $\left(0,\frac{5b-3}{2},\frac{5-3a}{2}\right)=\left(0,\frac{17}{2},-\frac{13}{2}\right)$

Equate the corresponding coordinates, we get  $\overline{)a=6,b=4}$  +91

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