Finding the area of a triangle is straightforward if you know the length of the base and the height of the triangle. but is it possible to find the area of a triangle if you know only the coordinates of its vertices? in this task, you’ll find out. consider δabc, whose vertices are a(2, 1), b(3, 3), and c(1, 6); let line segment ac represent the base of the triangle.

part a
find the equation of the line passing through b and perpendicular to ac

part b
let the point of intersection of ac with the line you found in part a be point d. find the coordinates of point d.

part c
use the distance formula to find the length of the base and the height of δabc.

part d
find the area of δabc.

PART A

We first need to work out the slope of AC
The slope of a line is given by vertical distance ÷ horizontal distance

A(2, 1) and C(1, 6)
Vertical distance = 
Horizontal distance = 
Slope of AC = 5 ÷ -1 = -5

A line that is perpendicular to AC will have slope -1 ÷ -5 = ¹/₅
We have the coordinate that this perpendicular line passes through, B(3, 3)
Use the formula , and we have ,  and 

 ⇒ This the required equation


PART B

To find the intersection point, we need to find the equation of the line that passes point A and point C
We have the slope from working out in PART A, 
We can choose either point A or point C to be (x₁, y₁) to be substituted into the formula to form the equation

Take A(2, 1)

We will find solution for y = -5x + 11 and y = ¹/₅ x + ¹²/₅
We can start by equating both equations

y = y
-5x + 11 = ¹/₅ x + ¹²/₅
-5x - ¹/₅ x = ¹²/₅ - 11
-²⁶/₅ x = -⁴³/₅
x = -⁴³/₅ ÷ -²⁶/₅
x = ⁴³/₂₆

Then substitute x = ⁴³/₂₆ into either  y = -5x + 11 or y = ¹/₅ x + ¹²/₅
y = ¹/₅ (⁴³/₂₆) + ¹²/₅ = ⁷¹/₂₆

So, the coordinate of intersection is (⁴³/₂₆, ⁷¹/₂₆)


PART C

Refer to the diagram below
The base of the triangle is AC and the height is BD

The length of AC is given by 
The length of BD is given by 



PART D 

The area of the triangle is given by ¹/₂ × height × base

Area = ¹/₂ × √26 × (7√26)/√26
Area = √26

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