Question

The value(s) of $\lambda$, for which the triangle with vertices $(6,10,10),(1,0,-5)$ and $(6,-10,\lambda )$ will be a right angled triangle is/ are

• A. $1$
• B. $\frac{70}{3},0$
• C. $35$
• D. $0,-\frac{70}{3}$

Answer 1

Answer: (B)

$\frac{70}{3},0$

Let the given points be $A,B$ and $C$ respectively.
Then, ${AB}^2=350,{AC}^2=500\lambda -20\lambda +{\lambda }^2{BC}^2=150+10\lambda +{\lambda }^2$
Now, ${AB}^2+{AC}^2={BC}^2\Rightarrow 350+500-20\lambda +{\lambda }^2=150+10\lambda +{\lambda }^2\Rightarrow \lambda =\frac{70}{3}$
Next, ${AB}^2+B{C}^2={AC}^2\Rightarrow 250+150+10\lambda +{\lambda }^2=500-20\lambda +{\lambda }^2\Rightarrow \lambda =0$
Further, ${BC}^2+A{C}^2={AB}^2\Rightarrow 150+10\lambda +12+500-20\lambda +{\lambda }^2=350\Rightarrow {\lambda }^2-5\lambda +150=0$,
which has no real solution.
$\therefore$ the triangle is right angle for $\lambda =0,\frac{70}{3}$