Question

the area of a triangle whose vertices are (-2,-2), (-1,-3), and (x,0) is 3 square units. Find the value of x​

Answer 1

$\huge\underline\mathrm{GivEn:-}$

The vertices of a triangle = (-2,-2), (-1,-3), and (x,0)

The area of a triangle = 3square units

$\huge\underline\mathrm{To\:Find:-}$

Find the value of 'x'

$\huge\underline\mathrm{SoLution:-}$

We have given that the area of triangle is 3 sq. units

According to Question :-

$x_{1} = - 2 \: \: \: \: \: \: \: \: y_{1} = - 1\\ x_{2} = - 1 \: \: \: \: \: \: \: \: y _{2} = - 3 \\ x _{3} = x \: \: \: \: \: \: \: \: \: \: \: y_{3} = 0$

To find the area of Triangle, we use the formula⤵

$\frac{1}{2} [x _{1}(y _{2} - y _{3}) + x _{2}(y _{3} - y _{1}) + x _{3}(y _{1} - y _{2})]$

Put the given values in the formula

$\frac{1}{2} [- 2( - 3 - 0)+(- 1)(0 - ( - 2)+ x( - 2 - ( - 3)]=\:3$

$\frac{1}{2}[-2(-3-0)-1(0+2)+x(-2+3)]=\:3$

$\frac{1}{2} [( - 2 \times - 3)( - 1 \times 2)(x \times 1)]=\:3$

$\frac{1}{2} ( 6 \times - 2 \times x )= 3$

$- 6x = 3$

$x = - 2$

Thus, the value of 'x' is -2

$\huge\underline\mathrm{ThAnKs...}$

Answer 2

Answer:

$\implies$ x = -2

$\implies$The value of 'x' is -2.

$\large\underline\mathrm{Given:-}$

• The vertices of a triangle (-2, -1) (-1, 3), and (x, 0)
• The area of a triangle = 3 sq. unit's

$\large\underline\mathrm{To \: find}$

• The value of 'x'.

$\large\underline\mathrm{Solution}$

• we have given that the area of triangle is 3 sq. unit's.

$\implies$ x1 = -2

$\implies$ x2 = -1

$\implies$ x3 = x

$\implies$ y1 = -1

$\implies$ y2 = -3

$\implies$ y3 = 0

The using the formula.

$\implies$ 1/2[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 3

$\implies$ 1/2[-2(-3 - 0) + (-1)(0 - (-2)) + x(-2 - (-3))] = 3

$\implies$ 1/2[-2(-3 - 0) + (-1)(0 + 2) + x(-2 + 3)] = 3

$\implies$ 1/2[-2 × -3)(-1 × 2)(x × 1)] = 3

$\implies$$\implies$ (6 × -2 × x) = 3

$\implies$ -6x = 3

$\implies$ x = -2

Thus,

The value of 'x' is -2.