If the line passing through (5, 1, a) and (3, | vedclass

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Question

(D) The equation of the line passing through $(5, 1, a)$ and $(3, b, 1)$ is given by: $\frac{x - 5}{5 - 3} = \frac{y - 1}{1 - b} = \frac{z - a}{a - 1}

Vedclass · Accepted Answer

$a = 6, b = 4$

Vedclass · Answer

(D) The equation of the line passing through $(5, 1, a)$ and $(3, b, 1)$ is given by: $\frac{x - 5}{5 - 3} = \frac{y - 1}{1 - b} = \frac{z - a}{a - 1}$ $\frac{x - 5}{2} = \frac{y - 1}{1 - b} = \frac{z - a}{a - 1}$ Since the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$ lies on this line,we substitute these coordinates into the equation: $\frac{0 - 5}{2} = \frac{\frac{17}{2} - 1}{1 - b} = \frac{\frac{-13}{2} - a}{a - 1}$ $-\frac{5}{2} = \frac{15}{2(1 - b)} = \frac{-13 - 2a}{2(a - 1)}$ From the first equality: $-\frac{5}{2} = \frac{15}{2(1 - b)} \implies 1 - b = -3 \implies b = 4$ From the second equality: $-\frac{5}{2} = \frac{-13 - 2a}{2(a - 1)} \implies -5(a - 1) = -13 - 2a -5a + 5 = -13 - 2a 3a = 18 \implies a = 6$ Thus,$a = 6$ and $b = 4$.

If the line passing through $(5, 1, a)$ and $(3, b, 1)$ intersects the $yz$-plane at $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$,then find the values of $a$ and $b$.

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