Question

The vertices of a triangle ABC are A(2, 3, 1), B(-2, 2, 0) and C(0, 1, -1).Find the magnitude of the line joining mid points of the sides AC and BC.

• A. $\frac{1}{\sqrt{2}}$ unit
• B. 1 unit
• C. $\frac{3}{\sqrt{2}}$ unit
• D. 2 unit

Answer 1

The correct option is C $\frac{3}{\sqrt{2}}$ unit

For a point $(h,k,l)$ dividing a line joining 2 points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ by the ratio $m:n$

$h=\frac{m\times x_1+n\times {x_2}^{{}^{\prime }}}{m+n}k=\frac{m\times y_1+n\times {y_2}^{{}^{\prime }}}{m+n}l=\frac{m\times z_1+n\times z_2}{m+n}$

So, mid-point of line $AC$ is

$x^{}=\frac{1\times 0+1\times 2}{2},y=\frac{1\times 1+3\times 1}{2},z=\frac{-1\times 1+1\times 1}{2}$

$x=1y=2z=0$

Mid point of $BC$

$x^{{}^{\prime }}=\frac{1\times 0+(-2)\times 1}{2};y^{{}^{\prime }}=\frac{1\times 1+1\times 2}{2};z^{{}^{\prime }}=\frac{1\times (-1)+1\times 0}{2}$

$x^{{}^{\prime }}=-1,y^{{}^{\prime }}=1.5,z^{{}^{\prime }}=-.5$

So, the distance between the mid points is

$d=\sqrt{{(1+1)}^2+{(2-1.5)}^2+{(0+0.5)}^2}=\frac{3}{\sqrt{2}}$