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Find the area of the triangle whose vertices are $(4,3),$ $(-1,0),$ and $(2,-4)$.

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Hint: Take the respective points as$({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})$ and $({{x}_{3}},{{y}_{3}})$. Use the formula $\dfrac{1}{2}\left[ {{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]$ and simplify. You will get the answer.

Complete step-by-step answer:
A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices $A,$ $B,$ and $C$ is denoted $\Delta ABC$.
In Euclidean geometry, any three points, when non-collinear, determine a unique triangle and simultaneously, a unique plane (i.e. a two-dimensional Euclidean space). In other words, there is only one plane that contains that triangle, and every triangle is contained in some plane. If the entire geometry is only the Euclidean plane, there is only one plane and all triangles are contained in it; however, in higher-dimensional Euclidean spaces, this is no longer true. This article is about triangles in Euclidean geometry, and in particular, the Euclidean plane, except where otherwise noted.
A triangle is a $3$-sided polygon sometimes (but not very commonly) called the trigon. Every triangle has three sides and three angles, some of which may be the same.
The sides of a triangle are given special names in the case of a right triangle, with the side opposite the right angle being termed the hypotenuse and the other two sides being known as the legs. All triangles are convex and bicentric. That portion of the plane enclosed by the triangle is called the triangle interior, while the remainder is the exterior.
A triangle has three sides, three vertices, and three angles. The sum of the three interior angles of a triangle is always $180{}^\circ $. The sum of the length of two sides of a triangle is always greater than the length of the third side.

The area of a triangle is equal to half of the product of its base and height.
But if the three vertices of triangle are given such as $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})$and $({{x}_{3}},{{y}_{3}})$ then the area of triangle is $\dfrac{1}{2}\left[ {{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]$.

We are given the vertices of triangle $(4,3),$ $(-1,0),$and $(2,-4)$.
So area of triangle$\begin{align}
  & =\dfrac{1}{2}\left[ 4(0-(-4))+(-1)((-4)-3)+2(3-0) \right] \\
 & =\dfrac{1}{2}\left[ 4(4)-(-7)+2(3) \right] \\
 & =\dfrac{1}{2}\left[ 16+7+6 \right] \\
 & =\dfrac{1}{2}\left[ 29 \right] \\
\end{align}$

Area of triangle$=\dfrac{29}{2}units$

So we get the area of the triangle as $\dfrac{29}{2}units$.

Note: Read the question carefully. Don’t miss any term while simplifying. You must know the properties of triangles. Also, you must know the area of the triangle. While substituting the points in the formula of the area of the triangle, do not miss any of the numbers. Avoid the silly mistakes.
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