Triangle bounded by three lines

We find the triangle bounded by the three lines $y = x$, $y = -x + 4$ and $y = 0$.

Finding the vertices

Each vertex comes from solving a pair of the lines simultaneously. There are three ways to choose two lines out of three, and those give the three vertices.

Area

Taking the base along the $x$-axis gives length $4$, and the height is the $y$-coordinate $2$ of the vertex $(2, 2)$, so the area is $\dfrac{1}{2} \times 4 \times 2 = 4$. The coordinate formula for the area confirms it.

It is a right isosceles triangle

This is no ordinary triangle. The line $y = x$ has slope $1$ and $y = -x + 4$ has slope $-1$, and the product is $-1$. The two lines therefore meet at a right angle, and the vertex $(2, 2)$ is the right angle.

Moreover the distance from $(0, 0)$ to $(2, 2)$ and the distance from $(4, 0)$ to $(2, 2)$ are both $\sqrt{2^2 + 2^2} = 2\sqrt{2}$, so the triangle is right isosceles. Computing the area as $\dfrac{1}{2} \times 2\sqrt{2} \times 2\sqrt{2} = 4$ gives the same value again.

When no triangle is formed

Three lines do not always bound a triangle.

• No two of them may be parallel
• The three must not pass through a single point

Only when both conditions hold is there a triangle. If two lines are parallel, that pair yields no vertex. If all three share one point, the three vertices collapse onto it and the triangle degenerates.

Replacing $y = 0$ with $y = 2$, for instance, makes all three lines pass through $(2, 2)$, and nothing is enclosed. The large dots on the graph are the three vertices.