Question

The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right-angled triangle whose opposite vertex is $(7,2,4)$. Then which of the following is not the side of the triangle?

• A. $\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$
• B. $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$
• C. $\frac{x-7}{3}=\frac{y-2}{5}=\frac{z-4}{-1}$
• D. None of these

Answer 1

Answer: (C)

Given are vertex $A(7,2,4)$ and line
$\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8} $.
General point on above line $B \equiv(5 \lambda-6,3 \lambda-10,8 \lambda-14)$
Direction ratios of line $A B$ are $\langle 5 \lambda-13,3 \lambda-12,8 \lambda-18>$
Direction ratios of line $B C$ are $<5,3,8>$
Since angle between $A B$ and $B C$ is $\pi / 4$, we have
$\cos \frac{\pi}{4}=\frac{(5 \lambda-3) 5+3(3 \lambda-12)+8(8 \lambda-18)}{\sqrt{5^{2}+3^{2}+8^{2}} \cdot \sqrt{(5 \lambda-13)^{2}+(3 \lambda-12)^{2}+(8 \lambda-18)^{2}}}$
Squaring and solving, we get $\lambda=3,2$
Hence, equation of lines are $\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$
and $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$.