A line through a variable point creates a triangle; can we find its area?

A line through a variable point creates a triangle; can we find its area?

Question

Find

1. the equation of the line with gradient $-2$ which passes through the point $(2t,t)$,

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1. the values of $t$ for which the area enclosed by this line and the coordinate axes is $4$ square units.

Interactive diagram

Can we achieve an area of $4$ by varying $t$?

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Solution

Using the general formula $y-y_1=m(x-x_1)$, the equation is $y-t=-2(x-2t)$, which rearranges to give $y=-2x+5t$ or $2x+y=5t$.

Find

1. the values of $t$ for which the area enclosed by this line and the coordinate axes is $4$ square units.

We begin by finding the points where this line intersects the axes.

When $x=0$, $y=5t$, and when $y=0$, $x=\frac52 t$, so the points of intersection are $(0,5t)$ and $(\frac52 t,0)$. A sketch of the situation looks like this; we have drawn two sketches, one for the case $t>0$ and the other for the case $t<0$.

The region enclosed by the line and the coordinate axes is a right-angled triangle, so its area is given by $\text{area} = \frac{1}{2} \times \frac{5}{2}t \times 5t = \frac{25}{4}t^2.$

We are asked to find the values of $t$ for which this area equals $4$, so we need $\frac{25}{4}t^2 = 4.$ This gives $t^2=\frac{16}{25}$ and so $t = \pm \frac{4}{5}$.

We have not been quite precise in this argument. The area is $\frac{1}{2} \times \frac{5}{2}t \times 5t=\frac{25}{4}t^2$ when $t$ is positive, but when $t$ is negative, the side lengths (which must be positive) are $-\frac52 t$ and $-5t$, so the area is then $\frac{1}{2} \times \bigl(-\frac{5}{2}t\bigr) \times (-5t)$. However, this still multiplies to give $\frac{25}{4}t^2$, so our final answer is correct, as can also been seen by the symmetry of the situation.