Question

A pt. R with x coordinates 4 lies on the line segment joining the points P(2,-3,-4) and Q(8,0,10). Find the coordinates of the pt. R.

Answer 1

Before moving onto the answer, let's see:

$\sf{\underline{What\:is\:a\:Section\:Formula?}}$

The coordinates of the pt. R which divides the line segment joining 2 points $\sf{P(x_{1}, y_{1}, z_{1})}$ and $\sf{Q(x_{2}, y_{2}, z_{2})}$ internally in the ratio m:n are:

$\boxed{( \frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1} }{m + n}, \frac{mz_{2} + nz_{1} }{m + n} )}$


Now,

Suppose R divides PQ in the ratio k:1, then, by section formula, the coordinates of R:

$\boxed{R\:( \frac{8k + 2}{k + 1}\:, \frac{ - 3}{k + 1}\:, \frac{10k + 4}{k + 1} )}$


$\sf{\underline{Values:}}$

$\sf{P\:(2,-3,4)}$

$\sf{x_{1}= 2,\: y_{1} = - 3,\: z_{1} = 4}$

$\sf{Q\:(8,0,10)}$

$\sf{x_{2}= 8,\: y_{2} = 0,\: z_{2} = 10}$

$\boxed{m = k,\:n = 1}$

$\sf{\underline{Since:}}$

(i) $\sf{\underline{x\:co-ordinate\:of\:R\:is\:4:}}$

$\implies$ $\frac{8k + 2}{k + 1} = 4$

$\implies$ $8k + 2 = 4(k + 1)$

$\implies$ $8k - 4k = 4 - 2$

$\implies$ $4k = 2$

$\implies$ $k = \frac{2}{4}$

$\implies$ $k =$ $\boxed{\frac{1}{2}}$


(ii) $\sf{\underline{y\:co-ordinate\:of\:R:}}$

Putting $k = \frac{1}{2}$

$\implies$ $\frac{ - 3}{k + 1} = \frac{ - 3}{( \frac{1}{2} + 1) }$

$\implies$ $\frac{ - 3}{ \frac{3}{2} }$

$\implies$ $- 3 \times \frac{2}{3}$

$\implies$ $\boxed{- 2}$


(iii) $\sf{\underline{z\:co-ordinate\:of\:R:}}$

Putting $k = \frac{1}{2}$

$\implies$ $\frac{10k + 4}{k + 1} = \frac{10( \frac{1}{2}) + 4 }{ (\frac{1}{2}) + 1}$

$\implies$ $\frac{9}{ \frac{3}{2} }$

$\implies$ $9 \times \frac{2}{3}$

$\implies$ $3 \times 2$

$\implies$ $\boxed{6}$


$\sf{\underline{We\:know\:that:}}$

$x = 4$, $y = - 2$, $z = 6$


$\sf{\underline{Therefore:}}$

The coordinates of pt. R $(4, - 2, 6)$

Answer 2

A pt. R with x coordinates 4 lies on the line segment joining the points P(2,-3,-4) and Q(8,0,10). Find the coordinates of the pt. R.