Question

Co-ordinate Geometry
for what value of a the area of triangle formed by the points (2, 4), (6. a) and
|-1, 1) is 9 sq. units?​

Answer 1

Answer:

The value of a  = 14

Step-by-step explanation:

Given

(x₁, y₁) = (2, 4)

(x₂, y₂) = (6, a)

(x₃, y₃) = (-1, 1)

A = 9 sq.units

Area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is:

$A=\frac{\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|}{2}$

substituting the values (x₁, y₁) = (2, 4), (x₂, y₂) = (6, a), (x₃, y₃) = (-1, 1), A = 9 in th formula

$A=\frac{\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|}{2}$

$9=\frac{\left|2\left(a-1\right)+6\left(1-4\right)+-1\left(4-a\right)\right|}{2}$

Multiply both sides by 2

$\frac{2\left|2\left(a-1\right)+6\left(1-4\right)-1\left(4-a\right)\right|}{2}=9\cdot \:2$

simplify

$\left|2\left(a-1\right)+6\left(1-4\right)-1\left(4-a\right)\right|=18$

As the area is always positive.

so

$2\left(a-1\right)+6\left(1-4\right)-1\cdot \left(4-a\right)=18$

$2\left(a-1\right)-18-\left(4-a\right)=18$

Add 18 to both sides

$2\left(a-1\right)-18-\left(4-a\right)+18=18+18$

simplify

$2\left(a-1\right)-\left(4-a\right)=36$

$3a-6=36$

$3a=42$

Divide both sides by 3

$\frac{3a}{3}=\frac{42}{3}$

$a=14$

Thus, the value of a  = 14