Question

What is the area of a triangle whose vertices are X(1, 1), Y(3, -1), and Z(4,4)?

Answer 1

Answer:

Step-by-step explanation:

Note that in this diagram, point A represents point X, point B represents point Y, and point C represents point Z.

From the diagram, it appears as if $\overline{XZ} \perp \overline{XY}$. To determine if this is the case, we can find the slopes of both segments.

$m_{\overline{XZ}}=\frac{4-1}{4-1}=1\\m_{\overline{XY}}=\frac{-1-1}{3-1}=-1$

Since these slopes are negative reciprocals, we know that they are perpendicular.

This means we can use the formula $A=\frac{1}{2}bh$, where b is the base (XY) and h is the height (XZ).

• Note these are interchangeable.

Using the distance formula,

$XY=\sqrt{(3-1)^{2}+(-1-1)^{2}}=2\sqrt{2}\\XZ=\sqrt{(4-1)^{2}+(4-1)^{2}}=3\sqrt{2}$

This means the area is $\frac{1}{2}(2\sqrt{2})(3\sqrt{2})=\boxed{6}$