Question

find the value of x, the area the triangle formed by the vertices (-2, -2),(-1, -3) and (x,0) is 3 sq. units​

Answer 1

Answer:

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Answer 2

Recall finding the area of triangle:

If it is a right-angled triangle and we know the base and height:

• $\text{Area of Triangle} = \dfrac{1}{2} \times \text{base} \times \text{height}$

If we know two of the lengths and the angle in-between:

• $\text{Area of Triangle} = \dfrac{1}{2} ab\sin C$

if we know all the three lengths of the triangle:

• $\text{Area of Triangle} = \sqrt{s(s - a)(s - b)(s - c)}$

If we know the three vertices only:

• $\text{Area of Triangle} = \dfrac{1}{2} (x1(y2- y3)+x2(y3 - y1)+x3(y1 -y2)$

Given the points of the 3 vertices:

Point 1 = (-2 , -2 )

Point 2 = (-1, -3)

Point 3 = (x, 0)

Find the area in term of x:

$\text{Area of Triangle} = \dfrac{1}{2} (x1(y2- y3)+x2(y3 - y1)+x3(y1 -y2)$

$\text{Area of Triangle} = \dfrac{1}{2} ((-2) (-3 - 0)+(-1)(0 - (-2))+(x)(-2 - (-3)))$

$\text{Area of Triangle} = \dfrac{1}{2} ((-2) (-3)+(-1)(2)+(x)(1))$

$\text{Area of Triangle} = \dfrac{1}{2} (6-2+x)$

$\text{Area of Triangle} = \dfrac{1}{2} (4+x)$

Given that the area is 3 units²:

$\dfrac{1}{2} (4+x) = 3$

$4 + x = 6$

$x = 6 - 4$

$x = 2$

Answer: The value of x is 2.