The co-ordinates of the point where the line through $\mathrm{A}(3,4,1)$ and $\mathrm{B}(5,1,6)$ crosses the…
- $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$
- $\left(-\frac{13}{5}, \frac{23}{5}, 0\right)$
- $\left(\frac{13}{5},-\frac{23}{5}, 0\right)$
- $\left(-\frac{13}{5},-\frac{23}{5}, 0\right)$
Solution
The equation of line passing through the points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$ is given by $\begin{array}{ll} & \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} \\ \therefore & \frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1} \\ \therefore & \frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5} \end{array}$
Since the line crosses the XY plane, $\mathrm{z}=0$ $\therefore \quad \frac{x-3}{2}=\frac{y-4}{-3}=\frac{-1}{5}$ $\therefore \quad \frac{x-3}{2}=\frac{-1}{5}$ and $\frac{y-4}{-3}=\frac{-1}{5}$ $\Rightarrow x=\frac{13}{5}$ and $y=\frac{23}{5}$ Required point is $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$
Asked in: MHT CET 2024 (04 May Shift 2)