The coordinates of the point where the line p | vedclass

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(A) Let $A(x_1, y_1, z_1) = (3, 4, 1)$ and $B(x_2, y_2, z_2) = (5, 1, 6)$.The equation of the line passing through two points is $\frac{x-x_1}{x_2-x_1

Vedclass · Accepted Answer

$\left(\frac{13}{5}, \frac{23}{5}, 0\right)$

Vedclass · Answer

(A) Let $A(x_1, y_1, z_1) = (3, 4, 1)$ and $B(x_2, y_2, z_2) = (5, 1, 6)$.The equation of the line passing through two points is $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.Substituting the values,we get $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1}$,which simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = k$.Since the line crosses the $xy$-plane,the $z$-coordinate must be $0$.Setting $z = 0$,we have $\frac{0-1}{5} = k$,so $k = -\frac{1}{5}$.Now,find $x$ and $y$ using $k = -\frac{1}{5}$:$x - 3 = 2k \Rightarrow x = 3 + 2(-\frac{1}{5}) = 3 - \frac{2}{5} = \frac{13}{5}$.$y - 4 = -3k \Rightarrow y = 4 - 3(-\frac{1}{5}) = 4 + \frac{3}{5} = \frac{23}{5}$.Thus,the required point is $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$.

The coordinates of the point where the line passing through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $xy$-plane are

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