A line cuts the axes to make a triangle; can we find its area?

A line cuts the axes to make a triangle; can we find its area?

Question

Three points have coordinates $A(-2, -1)$, $B(6, 9)$ and $C(2, -3)$. The line through the midpoint of $AB$ parallel to $AC$ meets the $x$-axis at $X$ and the $y$-axis at $Y$. Calculate the coordinates of $X$ and $Y$. Hence deduce the area of $\triangle XOY$, where $O$ is the origin.

Suggestion

It is always helpful to draw a sketch of a geometric situation.

We are given coordinates of points; how can we use these to find the equation of the new line?

Solution

We will first find the midpoint of $AB$; let’s call it $M$.

The midpoint of $AB$ has coordinates which are midway between the coordinates of $A$ and $B$.

So in general, given two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the midpoint has coordinates $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$.

In this case, therefore, $M$ has coordinates $\left(\dfrac{-2 + 6}{2}, \dfrac{-1 + 9}{2}\right) = (2, 4)$.

Another method

We can consider this sketch of the situation:

Now the horizontal distance between $A$ and $B$ is $6 - (-2) = 8$ and the vertical distance between $A$ and $B$ is $9 - (-1) = 10$.

Therefore, the horizontal distance between $A$ and $M$ is $\dfrac{8}{2} = 4$ and the vertical distance between $A$ and $M$ is $\dfrac{10}{2} = 5$.

As the point $A$ has the coordinates $(-2, -1)$, the midpoint $M$ is $(-2 + 4, -1 + 5) = (2,4)$.

What is the equation of the line through $M$ parallel to $AC$? We will use the fact that two lines are parallel if and only if they have the same gradient.

The gradient of the line $AC$ is $\dfrac{-3-(-1)}{2-(-2)}=\dfrac{-2}{4} = - \dfrac{1}{2}$. So the line through $M$ parallel to $AC$ has gradient $-\dfrac12$.

As it passes through $M(2,4)$, it therefore has equation $y-4=-\frac12(x-2)$ (using the general formula $y-y_1=m(x-x_1)$), which rearranges to $y=-\frac12 x+5$.

We now want to find $X$ and $Y$, which are the intercepts with the $x$ and $y$ axes respectively.

To find $X(x,0)$, we solve the equation $0 = -\dfrac{1}{2} x + 5$, and therefore we have $X=(10, 0)$.

To find $Y(0,y)$, we solve $y = -\dfrac{1}{2} \times 0 + 5$, which gives $y = 5$. Therefore, $Y$ has coordinates $(5, 0)$.

The triangle $\triangle XOY$ is right-angled, so its area is given by $\frac12 OX\times OY=\frac 12\times 10\times 5=25$.