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(A) The equation of the line passing through points $A(3, 5, -1)$ and $B(6, 3, -2)$ is given by the formula $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-

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$-\frac{5}{2}$

Vedclass · Answer

(A) The equation of the line passing through points $A(3, 5, -1)$ and $B(6, 3, -2)$ is given by the formula $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.Substituting the coordinates,we get $\frac{x-3}{6-3} = \frac{y-5}{3-5} = \frac{z-(-1)}{-2-(-1)}$,which simplifies to $\frac{x-3}{3} = \frac{y-5}{-2} = \frac{z+1}{-1} = r$.Any point $P$ on this line can be represented as $(3r+3, -2r+5, -r-1)$.Given that the $y$-coordinate of point $P$ is $2$,we set $-2r+5 = 2$.Solving for $r$,we get $-2r = -3$,so $r = \frac{3}{2}$.Now,substitute $r = \frac{3}{2}$ into the $z$-coordinate expression: $z = -r-1 = -\frac{3}{2} - 1 = -\frac{5}{2}$.

If a point $P$ on the line segment joining the points $(3, 5, -1)$ and $(6, 3, -2)$ has its $y$-coordinate $2$,then its $z$-coordinate is

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