Question

Let R be the point which divides the line segment joining P(2,1,0) and Q(-1,3,4) in the ratio 1:2 such that PR<PQ. Find the =n of the line passing through R and parallel to the line x/2=y/1=z/3

Answer 1

Answer on Question #46834 – Math – Analytic Geometry

Problem.

Let $R$ be the point which divides the line segment joining $P(2,1,0)$ and $Q(-1,3,4)$ in the ratio 1:2 such that $PR < PQ$. Find the $=n$ of the line passing through $R$ and parallel to the line $x/2 = y/1 = z/3$

Solution:

Let $R$ has coordinates $(a, b, c)$. Then $2\overline{PR} = \overline{RQ}$. $\overline{PR} = (a - 2, b - 1, c)$ and $\overline{RQ} = (-1 - a, 3 - b, 4 - c)$. Hence $2(a - 2, b - 1, c) = (-1 - a, 3 - b, 4 - c)$. Therefore $2a - 4 = -1 - a$, $2b - 2 = 3 - b$, $2c = 4 - c$. Hence $a = 1$, $b = \frac{1}{3}$, $c = \frac{4}{3}$, $R\left(1, \frac{1}{3}, \frac{4}{3}\right)$. The line, that passes through $R\left(1, \frac{1}{3}, \frac{4}{3}\right)$ and is parallel to the line $x/2 = y/1 = z/3$, has equation

$$\frac{x - 1}{2} = \frac{y - \frac{1}{3}}{1} = \frac{z - \frac{4}{3}}{3}$$

or

$$\frac{x - 1}{2} = \frac{3y - 1}{3} = \frac{3z - 4}{9}.$$

Answer: $\frac{x - 1}{2} = \frac{3y - 1}{3} = \frac{3z - 4}{9}$.

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