Ans: (a) (i) Parallel (ii) $x-3y-6=0$ (b) (i) Yes (ii) $1:1$
-
- $\Gamma$ and $L$ are parallel to each other.
- Note that the $y$ intercept of $\Gamma$ is $-2$. Since $\Gamma$ is parallel to $L$, then we have
$\begin{array}{cl}
& \text{The slope of }\Gamma \\
= & \text{The slope of }L \\
= & \dfrac{0-(-1)}{3-0} \\
= & \dfrac{1}{3}
\end{array}$Therefore, the equation of $\Gamma$ is $y = \dfrac{1}{3}x -2$.
-
- Note that $Q=(6,0)$. For $x=6$, we have
$\begin{array}{rcl}
y & = & \dfrac{1}{3}(6)-2 \\
& = & 0
\end{array}$Therefore $\Gamma$ passes through $Q$.
- By (b)(i), $\Gamma$ passes through $Q$. Then $HQ$ and $KQ$ are radii of the circle $C$. Note that $\Gamma$ is parallel to $L$. Then the perpendicular distances from $A$ to $HQ$ and from $B$ to $KQ$ are equal. Therefore the required ratio is $1:1$.
- Note that $Q=(6,0)$. For $x=6$, we have
