Question

The equation of the line passing through the $(-2,1,0)$ and $(3,4,-1)$ are

• A. $\frac{x-3}{5}=\frac{y-4}{3}=\frac{z+1}{-1}$
• B. $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z}{5}$
• C. $\frac{x+3}{5}=\frac{y+4}{3}=\frac{z-1}{-1}$
• D. $\frac{x+2}{5}=\frac{y-2}{3}=\frac{z}{-1}$

Answer 1

Answer: (A)

$\frac{x-3}{5}=\frac{y-4}{3}=\frac{z+1}{-1}$

In the equation of the plane of the form $ax+by+cz$, then, b and c represent the direction ratios of normal vector to the plane.

The general equation of a line passing through points $(x_1,y_1,z_1)$ & $(x_2,y_2,z_2)$is

$\frac{(x-x_1)}{(x_2-x_1)}=\frac{(y-y_1)}{(y_2-y_1)}=\frac{(z-z_1)}{(z_2-z_1)}=k,say$...................where $k$ is any constant

The equation of Line passing through the point $(-2,1,0)$ and $(3,4,-1)$

$\frac{(x-3)}{(3+2)}=\frac{(y-4)}{(4-1)}=\frac{(z+1)}{(-1+0)}$

$⟹\frac{x-3}{5}=\frac{y-4}{3}=\frac{z+1}{-1}$