A straight line passes through the points $P \left(-1, 4\right)$ and $Q \left(5, -2\right)$. It intersects the coordinate axes at points $A$ and $B$. $M$ is the midpoint of segment $AB$. Find:
- the equation of the line;
- the coordinates of points $A$ and $B$;
- the coordinates of $M$.
- Given: $\;$ $P \left(x_1, y_1\right) \equiv \left(-1, 4\right)$, $\;$ $Q \left(x_2, y_2\right) \equiv \left(5, -2\right)$
Slope of line $PQ = m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-2 - 4}{5 - \left(-1\right)} = \dfrac{-6}{6} = -1$
$\therefore \;$ Equation of line $PQ$ is: $\;$ $y - y_1 = m \left(x - x_1\right)$
i.e. $\;$ $y - 4 = -1 \left(x + 1\right)$
i.e. $\;$ $y - 4 = -x - 1$
i.e. $\;$ $x + y = 3$ - Let the line $PQ$ intersect the X-axis at the point $A$ and the Y-axis at the point $B$.
Then, the coordinates of point $A$ are $\left(a, 0\right)$ and
the coordinates of point $B$ are $\left(0, b\right)$.
When the line $PQ$ cuts the X-axis, putting $y = 0$ in the equation of $PQ$ gives $\;$ $x = 3$
When the line $PQ$ cuts the Y-axis, putting $x = 0$ in the equation of $PQ$ gives $\;$ $y = 3$
$\therefore \;$ $A = \left(3, 0\right)$; $\;$ $B = \left(0, 3\right)$ - $M$ is the midpoint of $AB$.
$\therefore \;$ Coordinates of $M = \left(\dfrac{3 + 0}{2}, \dfrac{0 + 3}{2}\right)$
i.e. $\;$ $M = \left(\dfrac{3}{2}, \dfrac{3}{2}\right)$