The equation of the straight line passing thr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{

Vedclass · Accepted Answer

$\frac{x - 4}{1} = \frac{y + 5}{-2} = \frac{z + 2}{-1}$

Vedclass · Answer

(A) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$.Given points are $(x_1, y_1, z_1) = (4, -5, -2)$ and $(x_2, y_2, z_2) = (-1, 5, 3)$.Substituting these values into the formula:$\frac{x - 4}{-1 - 4} = \frac{y - (-5)}{5 - (-5)} = \frac{z - (-2)}{3 - (-2)}$Simplifying the denominators:$\frac{x - 4}{-5} = \frac{y + 5}{10} = \frac{z + 2}{5}$Dividing the denominators by $-5$:$\frac{x - 4}{1} = \frac{y + 5}{-2} = \frac{z + 2}{-1}$.Thus,the correct option is $A$.

The equation of the straight line passing through the points $(4, -5, -2)$ and $(-1, 5, 3)$ is

Similar Questions

Show that the points $A(1, 2, 7)$,$B(2, 6, 3)$,and $C(3, 10, -1)$ are collinear.

The length of the perpendicular from the point $A(1, -2, -3)$ on the line $\frac{x-1}{2} = \frac{y+3}{-1} = \frac{z+1}{-2}$ is (in $\text{ units}$)

The distance between parallel lines $\vec{r}=(2\hat{i}-\hat{j}+\hat{k})+\lambda(2\hat{i}+\hat{j}-2\hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(2\hat{i}+\hat{j}-2\hat{k})$ is

The length of the perpendicular drawn from the point $P(5, 4, -1)$ to the line $\frac{x - 1}{2} = \frac{y}{9} = \frac{z}{5}$ is:

If $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$ are points in a plane, then the vector equation of the line passing through the centroids of $\triangle ABD$ and $\triangle ACD$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module