Line L passes through two points (2, -3, 1) a | vedclass

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(D) 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_

Vedclass · Accepted Answer

$12$

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(D) 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 $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.Substituting the given points $(2, -3, 1)$ and $(3, -4, -5)$:$\frac{x-2}{3-2} = \frac{y-(-3)}{-4-(-3)} = \frac{z-1}{-5-1}$$\frac{x-2}{1} = \frac{y+3}{-1} = \frac{z-1}{-6} = k$ (let).For the point $(0, a, b)$ to lie on the line,we set $x = 0$:$\frac{0-2}{1} = k \implies k = -2$.Now,find $a$ and $b$ using $k = -2$:$\frac{a+3}{-1} = -2 \implies a+3 = 2 \implies a = -1$.$\frac{b-1}{-6} = -2 \implies b-1 = 12 \implies b = 13$.Therefore,$a+b = -1 + 13 = 12$.

Line $L$ passes through two points $(2, -3, 1)$ and $(3, -4, -5)$. If point $(0, a, b)$ lies on the line $L$,then $a+b =$ . . . . . . .

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