Answer:
The area of ΔABC is 3.5 unit²
Step-by-step explanation:
Given that the coordinates of the triangle are;
A = (2, 1), B = (3, 3), C = (1, 6)
The slope, m₁ of the line joining A to C is found as follows;
m₁ = (6 - 1)/(1 - 2) = -5
From the general equation of a straight line, y =m·x + c, we have;
6 = -5×1 + c
∴ c = 6 + 5 = 11
The equation of the line AC is therefore;
y = -5·x + 11
The equation of a perpendicular line to another line is the negative reciprocal of the slope of the line
Therefore, we have;
y - 3 = 1/5(x - 3)
y = 1/5x - 3/5 + 3
y = 1/5x + 12/5
5·y = x + 12
At the point of intersection, D, of the line from vertex B and the line AC, we have;
Equating both lines to each other gives;
y = -5·x + 11 = 1/5x + 12/5
1/5x + 5·x = 11 - 12/5 = 42/5
26/5·x = 42/5
x = 43/26
y = 71/26
D = (43/26, 71/26)
The length of the base AC = √((2 - 1)² + (1 - 6)²) = √26
The height of ΔABC = length of segment BD = √((43/26 - 3)²+ (71/26 - 3)²) = 7/√26
The area of triangle ABC = 1/2×AC×BD = 1/2×√26×7/√26 = 7/2 = 3.5 unit²
The area of ΔABC = 3.5 unit².
