On the line (x-8)/(8)=(y-2)/(-6)=tfrac z0 find the point…

On the line $\tfrac{x-8}{8}=\tfrac{y-2}{-6}=\tfrac z0$ find the point whose distance from $(8,2,0)$ is $10$.

Explanation of Method

The points of a line at distance 10 from the point (8, 2, 0).

Step 1: Solution

The fraction $\tfrac z0$ means $z=0$. Parametric form: $x=8+8t,\ y=2-6t,\ z=0$. The point $(8,2,0)$ is on the line ($t=0$), and $|\vec d|=10$, so the distance is $10|t|$.

\[t=\pm1:\qquad P_1(16,-4,0),\qquad P_2(0,8,0)\]

Check: $\sqrt{64+36}=10$ in both cases.

$A(8,2)$ $P_1(16,-4)$ $P_2(0,8)$

The line lies in the plane $z=0$ (view from above). The circle of radius $10$ about $A(8,2,0)$ meets it at $P_1$ and $P_2$.

Result

\[\boxed{\,t=\pm1:\qquad P_1(16,-4,0),\qquad P_2(0,8,0)\,}\]