Question

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

• A. $a=8,b=2$
• B. $a=2,b=8$
• C. $a=4,b=6$
• D. $a=6,b=4$

Answer 1

The correct option is D

$a=6,b=4$

Explanation for the correct option

Given points are $\left(5,1,a\right)$ and $\left(3,b,1\right)$

Equation of the lines is given by, $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Thus, $\frac{x-5}{3-5}=\frac{y-1}{b-1}=\frac{z-a}{1-a}$

It is given that $\left(0,\frac{17}{2},\frac{-13}{2}\right)$ satisfies the equation, so, replacing these values for $x,y,z$, we get,

$\Rightarrow \frac{0-5}{3-5}=\frac{{\displaystyle \frac{17}{2}}-1}{b-1}=\frac{{\displaystyle \frac{-13}{2}}-a}{1-a}$

Let, $\frac{5}{2}=\frac{{\displaystyle \frac{17}{2}}-1}{b-1}\Rightarrow b-1=\frac{15}{5}\Rightarrow b=3+1$, so, $b=4$

Also, $\frac{5}{2}=\frac{{\displaystyle \frac{-13}{2}}-a}{1-a}\Rightarrow 1-a=-\frac{13}{5}-\frac{2a}{5}\Rightarrow a=\frac{5}{3}\times \frac{18}{5}$, so, $a=6$

Therefore, option (D) i.e. $a=6,b=4$ is the correct answer.