Area - Triangle, Given Three Vertices, 3

Category: Analytic Geometry, Plane Geometry

"Published in Newark, California, USA"

Show that the points (5, 4), (-2, 1), and (2, -3) are the vertices of an isosceles triangle, and find its area.

Solution: 

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

The first thing that we need to do is to get the length of the sides of a triangle using the distance of two points as follows

For the length of AB:

For the length of AC:

For the length of BC:

Since AB ≅ AC, then the given triangle is an isosceles triangle. 

The area of a triangle is given by the formula

Substitute the values of the coordinates of the vertices of a triangle to the above equation and solve for the value of matrix or determinant, we have

Therefore, the area of a triangle is

 

Area - Triangle, Given Three Vertices, 2

Category: Analytic Geometry, Plane Geometry

"Published in Newark, California, USA"

Show that the points (-2, 0), (2, 3), and (5, -1) are the vertices of a right triangle, and find its area.

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

The first thing that we need to do is to get the slope of each sides of a triangle using the two point formula as follows

For the slope of AB:

For the slope of BC:

For the slope of AC:

Since the slope of AB is negative reciprocal of the slope of BC, then the given triangle is a right triangle. A right triangle is a triangle whose two sides are perpendicular . 

The area of a triangle is given by the formula

Substitute the values of the coordinates of the vertices of a triangle to the above equation and solve for the value of matrix or determinant, we have

The area of any closed plane figure is always in absolute value. In this case, if the calculated area is negative, then change the negative sign into positive sign. Therefore, the area of a triangle is

Square, Rectangle, Parallelogram Problems, 3

Category: Analytic Geometry, Plane Geometry

"Published in Newark, California, USA"

Three vertices of a parallelogram are (1, 3), (0, 0), and (4, 0). Find the three possible locations of the fourth vertex.

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

There are three possible locations of the fourth vertex of a parallelogram. The two points or vertices can be used to draw one of the sides or one of the diagonals of a parallelogram.

Case 1: If you will use the three points or vertices to draw the two sides of a parallelogram, the fourth vertex is located at the first quadrant as shown in the figure

Photo by Math Principles in Everyday Life

Since one of the sides of a parallelogram is a horizontal line that lies along the x-axis, the fourth vertex of a parallelogram is P(1 + 4, 3) or P (5, 3).

Case 2: If you will use the three points or vertices to draw the two sides of a parallelogram, the fourth vertex is located at the second quadrant as shown in the figure

Photo by Math Principles in Everyday Life

Since one of the sides of a parallelogram is a horizontal line that lies along the x-axis, the fourth vertex of a parallelogram is P(1 - 4, 3) or P (-3, 3).

Case 3: If you will use the three points or vertices to draw the two diagonals of a parallelogram, the fourth vertex is located at the fourth quadrant as shown in the figure

Photo by Math Principles in Everyday Life

The center of a parallelogram which is the intersection of two diagonals bisects the diagonals into two equal parts. In this case, the midpoint of a diagonal that lies along the x-axis is (2, 0). If the midpoint of the other diagonal is also (2, 0), then we can solve for the coordinates of the other end of other diagonal which the fourth vertex of a parallelogram as follows

Therefore, the coordinates of the other end of other diagonal or the fourth vertex of a parallelogram is P(3, -3).