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Triangle with No Points at (0,0)

What is a Triangle with No Points at (0,0)(0, 0)?

In coordinate geometry, a triangle may have all its vertices away from the origin. The general approach to calculating properties such as area or side lengths remains similar but without the simplifications that arise when one vertex is at (0,0)(0, 0)

Area of the Triangle

The area of a triangle with vertices (x1,y1x_1, y_1), (x2,y2x_2, y_2), and (x3,y3x_3, y_3) is given by:

Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

This formula works for any triangle on the Cartesian plane and is derived from the determinant of a matrix that represents the vertices.

Lengths of the Sides

From (x1,y1x_1, y_1) to (x2,y2x_2, y_2):

Length=(x2x1)2+(y2y1)2\text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

From (x2,y2x_2, y_2) to (x3,y3x_3, y_3):

Length=(x3x2)2+(y3y2)2\text{Length} = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}

From (x3,y3x_3, y_3) to (x1,y1x_1, y_1):

Length=(x1x3)2+(y1y3)2\text{Length} = \sqrt{(x_1 - x_3)^2 + (y_1 - y_3)^2}

Worked Examples

infoNote

Example 1: Find the Area of a Triangle

Problem: Calculate the area of a triangle with vertices at (1,2),(4,6)(1, 2), (4, 6), and (5,3)(5, 3)


Solution:

Step 1: Using the area formula:

Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

Step 2: Substitute (x1,y1)=(1,2),(x2,y2)=(4,6),(x3,y3)=(5,3)(x_1, y_1) = (1, 2), (x_2, y_2) = (4, 6), (x_3, y_3) = (5, 3)

Area=121(63)+4(32)+5(26)\text{Area} = \frac{1}{2} \left| 1(6 - 3) + 4(3 - 2) + 5(2 - 6) \right|=121(3)+4(1)+5(4)=123+420= \frac{1}{2} \left| 1(3) + 4(1) + 5(-4) \right| = \frac{1}{2} \left| 3 + 4 - 20 \right| =1213=132= \frac{1}{2} \left| -13 \right| = \frac{13}{2}

Answer: The area is 6.56.5 square units.


infoNote

Example 2: Find the Length of a Side

Problem: Calculate the length of the side joining (2,3)(2, 3) and (6,7)(6, 7)


Solution:

Step 1: Using the distance formula:

Length=(x2x1)2+(y2y1)2\text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Step 2: Substitute (x1,y1)=(2,3),(x2,y2)=(6,7)(x_1, y_1) = (2, 3), (x_2, y_2) = (6, 7)

Length=(62)2+(73)2=42+42\text{Length} = \sqrt{(6 - 2)^2 + (7 - 3)^2} = \sqrt{4^2 + 4^2} =16+16=32=42= \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}

Answer: The length is 424\sqrt{2} units.


Summary

  • For triangles with no vertices at (0,0)(0, 0), the area formula is:
Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
  • The distance formula calculates the lengths of the sides.
  • This method applies universally to all triangles in the Cartesian plane.
  • Practice finding areas and side lengths to become proficient with these calculations.
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