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 find the equations of the remaining sides.

Answer 1

$\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}=\lambda$(say)
General point on above line is $(5\lambda -6,3\lambda -10,8\lambda -14).$
Let $B\equiv (5\lambda -6,3\lambda -10,8\lambda -14)$
Dr's of line $AB$ are $<(5\lambda -6-7),(3\lambda -10-2),(8\lambda -14-4)>i.e.,<5\lambda -13,3\lambda -12,8\lambda -18>$
Also,dr's of line $BC$ are $<5,3,8>$
Since angle between $AB$ and $BC$ is $\frac{\pi }{4}$
$\Rightarrow \cos \frac{\pi }{4}=\frac{|(5\lambda -13)5+3(3\lambda -12)+8(8\lambda -18)|}{\sqrt{5^2+3^2+8^2}\sqrt{{(5\lambda -13)}^2+{(3\lambda -12)}^2+{(8\lambda -18)}^2}}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{|25\lambda +9\lambda +64\lambda -65-36-144|}{\sqrt{98}\sqrt{{(5\lambda -13)}^2+{(3\lambda -12)}^2+{(8\lambda -18)}^2}}$
Solving above equation we get $\lambda =3,2$
Hence, equation of line are
$\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$ and $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$