Question

FIND The value of x, if the
area the triangle formed by
the vertices (-2,-2),(-1,-3)
& (X,o) is 3sq units .​

Answer 1

Solution :-

Let the coordinates be A(-2,-2) B(-1,-3) C(x,0)

Given,

Area of the ΔABC = 3 sq. units

We know that,

$\text{Area of the triangle} = \dfrac{1}{2} \mid x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \mid$

Here,

• x1 = - 2
• y1 = - 2
• x2 = - 1
• y2 = - 3
• x3 = x
• y3 = 0

By substituting the values

$\implies 3 = \dfrac{1}{2} \mid - 2( - 3- 0) - 1(0 + 2) + x( - 2 + 3) \mid$

$\implies 6= - 2( - 3) - 1(2) + x(1)$

$\implies 6= 6 -2 + x$

$\implies 6= 4 + x$

$\implies 6 - 4= x$

$\implies 2 = x$

$\implies x = 2$

Therefore the value of x is 2.

Answer 2

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➠ To Find :

We have to find the value of x.

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➠ Solution :

We know the formula to find the area of the triangle when vertices are given

$\large{\star{\boxed{\sf{Area = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2|}}}}$

Where,

• x1 = - 2
• x2 = - 1
• x3 = x
• y1 = -2
• y2 = -3
• y3 = 0

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(Putting Values)

$\sf{3 = \dfrac{1}{2} \mid - 2( - 3- 0) - 1(0 + 2) + x( - 2 + 3)}$

$\sf{\mapsto 6= - 2( - 3) - 1(2) + x(1)}$

$\sf{\mapsto 6= 6 -2 + x}$

$\sf{ \mapsto 6= 4 + x}$

$\sf{\mapsto x = 6 - 4}$

$\sf{\mapsto x = 2}$