2010-II-51

Ans: A
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

$\begin{array}{rcl}{\displaystyle \frac{x}{16}}+{\displaystyle \frac{y}{12}}& =& 1\\ 3x+4y& =& 48\\ 3x+4y–48& =& 0\end{array}$