Question

prove that the points (-2,3,5),(1,2,3) and (7,0,-1) are collinear

Answer 1

Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively.

Points P, Q, and R are collinear if they lie on a line.

Answer 2

The points (- 2, 3, 5), (1, 2, 3) and (7, 0, - 1) are collinear, proved.

Step-by-step explanation:

Let the three points  A(- 2, 3, 5), B(1, 2, 3) and C(7, 0, - 1)

Prove that the points (- 2, 3, 5), (1, 2, 3) and (7, 0, - 1) are collinear.

Using distance formula,

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

$AB=\sqrt{(1+2)^{2}+(2-3)^{2}+(3-5)^{2}}$

$=\sqrt{(3)^{2}+(-1)^{2}+(-2)^{2}}=\sqrt{9+1+4}$

=$\sqrt{14}$

$BC=\sqrt{(7-1)^{2}+(0-2)^{2}+(-1-3)^{2}}$

$=\sqrt{(6)^{2}+(-2)^{2}+(-4)^{2}}=\sqrt{36+4+16}$

$=\sqrt{56}=2\sqrt{14}$

$=2\sqrt{14}$

Also,

$CA=\sqrt{(7+2)^{2}+(0-3)^{2}+(-1-5)^{2}}$

$=\sqrt{(9)^{2}+(-3)^{2}+(-6)^{2}}=\sqrt{81+9+36}$

$=\sqrt{126}=\sqrt{9\times 14}$

$=3\sqrt{14}$

∴ AB + BC $=\sqrt{14}+2\sqrt{14}$

$=3\sqrt{14}$ = CA, proved.