Question

Find the equation of the line joining (-2,1,3) and (1,1,4)​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Points are (-2,1,3) and (1,1,4)}$

$\underline{\textbf{To find:}}$

$\textsf{The equation of the line joining (-2,1,3) and (1,1,4)}$

$\underline{\textbf{Solution:}}$

$\mathsf{The\;equation\;of\;the\;line\;joining\;(x_1,y_1,z_1)\;and\;(x_2,y_2,z_2)\;is}$

$\bf\;\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}$

$\mathsf{Here,}$

$\mathsf{(x_1,y_1,z_1)= (-2,1,3)}$

$\mathsf{(x_2,y_2,z_2)= (1,1,4)}$

$\implies\mathsf{\dfrac{x+2}{1+2}=\dfrac{y-1}{1-1}=\dfrac{z-3}{4-3}}$

$\implies\boxed{\bf\dfrac{x+2}{3}=\dfrac{y-1}{0}=\dfrac{z-3}{1}}$

$\textsf{which is the equation of the required line}$