Ans: A
Consider the equation of the straight line
Consider the equation of the straight line
$\begin{array}{rcl}
5x – 3y & = & 30 \\
y & = & \dfrac{5}{3} x – 10
\end{array}$
Therefore, the $y$-intercept of the straight line is $-10$. Hence the coordinates of $B$ are $(0, -10)$.
Sub. $y=0$ into $5x-3y=30$, we have
$\begin{array}{rcl}
5x – 3(0) & = & 30 \\
x & = & 6
\end{array}$
Therefore, the $x$-intercept of the straight line is $6$. Hence the coordinates of $A$ are $(6,0)$. Therefore, the coordinates of the mid-point of $AB$
$\begin{array}{cl}
= & \left( \dfrac{0+6}{2} , \dfrac{-10 + 0 }{2} \right) \\
= & (3,-5)
\end{array}$
