Question

In a triangle $ABC$, if the mid-points of sides $AB,BC,CA$ are $(3,0,0),(0,4,0),(0,0,5)$ respectively, then $A{B}^2+B{C}^2+C{A}^2=$

• A. 50
• B. 200
• C. 300
• D. 400

Answer 1

Answer: (D)

Given, mid-point of sides $AB,BC,CA$ are
$(3,0,0),(0,4,0),(0,0,5)$.

$\therefore A=(3,-4,5)$
$B=(3,4,-5)$ and $C=(-3,4,5)$
$A{B}^2={\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2$
$=(3-3{)}^2+(4+4{)}^2+(-5-5{)}^2$
$=8^2+10^2=164$
$B{C}^2=(-3-3{)}^2+(4-4{)}^2+(5+5{)}^2$
$=36+100=136$
And $C{A}^2=(-3-3{)}^2+(4+4{)}^2+(5-5{)}^2$
$=(-6{)}^2+(8{)}^2=36+64=100$
Now, $A{B}^2+B{C}^2+C{A}^2=164+136+100=400$