Question

If \((5,3),(4,2)\) and \((1,-2)\) are the mid points of sides of triangle \(\mathrm{ABC}\), then the area of \(\triangle \mathrm{ABC}\) is (1) 2 sq. units (2) \(3 \mathrm{sq}\). units (3) \(1 \mathrm{sq}\). units (4) \(4 \mathrm{sq}\). units

Short answer

Answer: The area of the triangle is 32 sq. units.

Step-by-step solution

Identify the given midpoints

We are given the midpoints of sides of the triangle ABC: \((5, 3)\), \((4, 2)\), and \((1, -2)\).

Label the midpoints and sides of the triangle

Let's label these midpoints as follows: Midpoint of side AB: \((5, 3)\) Midpoint of side BC: \((4, 2)\) Midpoint of side AC: \((1, -2)\)

Use midpoint formula to find the vertices of the triangle

Let the coordinates of vertices A, B, and C be \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\), respectively. Using the midpoint formula, we have: Midpoint of AB \(= \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1+y_2}{2}\right) = (5, 3)\) Midpoint of BC \(= \left(\dfrac{x_2 + x_3}{2}, \dfrac{y_2+y_3}{2}\right) = (4, 2)\) Midpoint of AC \(= \left(\dfrac{x_1 + x_3}{2}, \dfrac{y_1+y_3}{2}\right) = (1, -2)\)

Solve the system of equations to find the vertices

We now have the following system of equations: \(\dfrac{x_1 + x_2}{2} = 5\) \(\dfrac{y_1 + y_2}{2} = 3\) \(\dfrac{x_2 + x_3}{2} = 4\) \(\dfrac{y_2 + y_3}{2} = 2\) \(\dfrac{x_1 + x_3}{2} = 1\) \(\dfrac{y_1 + y_3}{2} = -2\) Solving this system, we find the coordinates of the vertices of the triangle: A: \((2, -4)\) B: \((8, 10)\) C: \((0, 2)\)

Calculate the area of the triangle using the cross product method

Let's use the cross product method to find the area of the triangle: Area of triangle ABC \(= \dfrac{1}{2}|(x_1y_2 + x_2y_3 + x_3y_1) - (y_1x_2 + y_2x_3 + y_3x_1)|\) Substituting the coordinates of vertices A, B, and C, we get: Area of triangle ABC \(= \dfrac{1}{2}|(2\cdot 10 + 8\cdot 2 + 0\cdot(-4)) - ((-4)\cdot 8 + 10\cdot 0 + 2\cdot 2)|\) Area of triangle ABC \(= \dfrac{1}{2}|(20 + 16) - (-32 + 4)|\) Area of triangle ABC \(= \dfrac{1}{2}|(36 - (-28))|\) Area of triangle ABC \(= \dfrac{1}{2}\cdot 64\) So, the area of triangle ABC is \(32\,\mathrm{sq}\text{.}\,\mathrm{units}\).

Key concepts

Coordinates of Vertices

Understanding how to find the coordinates of the vertices in a geometric shape is fundamental in geometry. When dealing with triangles, like in our exercise, the task may involve using the midpoint formula. This formula enables us to determine the unknown vertices of a triangle when the midpoints of its sides are known.

The midpoint of a line segment is simply the average of the x-coordinates and the y-coordinates of the endpoints. This is mathematically expressed as \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\), where \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of the endpoints. By knowing the coordinates of the midpoints — which in essence are the averages of the vertices coordinates — we can set up a system of equations that when solved, reveal the coordinates of the unknown vertices.

To illustrate, if the midpoint is \(5, 3\), and it is the midpoint of vertices A and B, we can write two equations where \(5 = \frac{x_1 + x_2}{2}\) and \(3 = \frac{y_1 + y_2}{2}\). These can be solved alongside other similar equations to find the precise values for the coordinates of each vertex.

Area of Triangle Calculation

Calculating the area of a triangle is a common problem in geometry with multiple methods available. One effective approach is using the determinant of a matrix formed by the coordinates of the vertices, which is essentially the cross product method mentioned in the solution.

In practical terms, if you're given the coordinates of the vertices — say A \(x_1, y_1\), B \(x_2, y_2\), and C \(x_3, y_3\) — you can plug them into the formula:\[\text{Area of } \triangle ABC = \dfrac{1}{2}\left|(x_1y_2 + x_2y_3 + x_3y_1) - (y_1x_2 + y_2x_3 + y_3x_1)\right|\]

This method is a straightforward computational way to finding the area by using the properties of determinants to give us a scalar area value. It’s important to note that the absolute value is taken to ensure the area is positive, as the determinant can yield a negative number depending on the orientation of the vertices.

In educational contexts, teaching this formula is helpful because it consolidates learning about both coordinate graphing and the properties of matrices, while providing a direct application to a common geometrical problem.

System of Equations in Geometry

When dealing with geometric problems, we frequently encounter a system of linear equations. This set of equations typically arises when we are given certain conditions — such as the coordinates of midpoints — and we need to find unknown quantities, like the vertices of a triangle.

To solve the system of equations, we use algebraic methods like substitution, elimination, or matrix operations. Each method has its own advantages and provides a clear step-by-step path to find the vertices' coordinates. In our exercise, after establishing equations based on the midpoint formula, solving the system reveals the location of each vertex.

A careful approach to solving these systems is critical since any error may lead to a larger mistake in calculating the area or other properties of the geometric figure. Thus, emphasis on checking the solution can greatly assist in mastering this part of geometry, leading to a better understanding of the relationship between equations and geometric figures.