Ans: A
Consider the coordinates of $A$. Sub. $y=0$ in the equation, we have
Consider the coordinates of $A$. Sub. $y=0$ in the equation, we have
$\begin{array}{rcl}
x^2 + (0)^2 -16x – 12(0) & = & 0 \\
x(x-16) & = & 0
\end{array}$
Therefore, $x=16$ or $x=0$ (rejected). Therefore $A=(16,0)$.
Consider the coordinates of $B$. Sub. $x=0$ in the equation, we have
$\begin{array}{rcl}
(0)^2 + y^2 -16(0) -12y & = & 0 \\
y(y-12) & = & 0
\end{array}$
Therefore, $y=12$ or $y=0$ (rejected). Therefore $B=(0,12)$.
Hence, the equation of the straight line passing through $A$ and $B$ is
$\begin{array}{rcl}
\dfrac{x}{16} + \dfrac{y}{12} & = & 1 \\
3x + 4y & = & 48 \\
3x + 4y – 48 & = & 0
\end{array}$
