Question

The coordinates of the point in which the line joining the points $(3,5,-7)$ and $(-2,1,8)$is intersected by the plane $yz$ are given by

• A. $\left(0,\frac{13}{5},2\right)$
• B. $\left(0,\frac{-13}{5},-2\right)$
• C. $\left(0,\frac{-13}{5},\frac{2}{5}\right)$
• D. $\left(0,\frac{13}{5},\frac{2}{5}\right)$

Answer 1

The correct option is A

$\left(0,\frac{13}{5},2\right)$

Explanation for correct option:

Step 1: Finding the equation of line

Given that the points are $(3,5,-7)$ and $(-2,1,8)$.

The equation of the line joining the two points will be

$\begin{array}{l}\frac{x-x_1}{x_1-x_2}=\frac{y-y_1}{y_1-y_2}=\frac{z-z_1}{z_1-z_2}\\ ⟹\frac{x-3}{3-(-2)}=\frac{y-5}{5-1}=\frac{z-(-7)}{(-7)-8}\\ ⟹\frac{x-3}{5}=\frac{y-5}{4}=\frac{z+7}{-15}\end{array}$

Now, the line joining the two points meets the $yz$ plane at $x=0$. Therefore,

$\begin{array}{l}\frac{0-3}{5}=\frac{y-5}{4}=\frac{z+7}{-15}\\ ⟹\frac{y-5}{4}=\frac{-3}{5}and\frac{z+7}{-15}=\frac{-3}{5}\end{array}$

Step 2: Finding the values of $y,z$:

$\begin{array}{l}\frac{y-5}{4}=\frac{-3}{5}\\ ⟹5(y-5)=-4\times 3\\ ⟹5y-25=-12\\ ⟹5y=13\\ ⟹y=\frac{13}{5}\end{array}$

$\begin{array}{l}\frac{z+7}{-15}=\frac{-3}{5}\\ ⟹5(z+7)=-3\times -15\\ ⟹5z+35=45\\ ⟹5z=10\\ ⟹z=\frac{10}{5}\\ ⟹z=2\end{array}$

Therefore the point $\left(0,y,z\right)$ is $\left(0,\frac{13}{5},2\right)$

Hence, option (A) is the correct answer