Question

If the area
of the AABC is 68 sq. units and the
vertices are A (6, 7) *(-4,1) and C (a, 9) taken in
order than find the value of a​

Answer 1

Given :

• Coordinates of A = ( 6 , 7 )

• Coordinates of B = ( - 4 , 1 )

• Coordinates of C = ( a , 9 )

• Area of triangle = 68 unit²

To Find :

• Value of a

Solution :

$\large \implies \boxed{\boxed{ \sf \green{Area_{\triangle} = \dfrac{1}{2} \bigg[x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2) \bigg]}}}$

Here

$\sf x_1 = 6 \: \: \: \: \: \: \: \: \: x_2 = - 4 \: \: \: \: \: \: \: x_3 = a \\ \\ \sf y_1 = 7 \: \: \: \: \: \: \: \: \: x_2 = 1 \: \: \: \: \: \: \: \: \: \: \: y_3 = 9$

Substitute values in formula

$\implies \sf 68 = \frac{1}{2} \bigg[6(1 - 9) - 4(9 - 7) + a(7 - 1) \bigg] \\ \\\implies \sf 68 = \frac{1}{2} \bigg[6( - 8) - 4(2) + a(6) \bigg] \\ \\\implies \sf 68 = \frac{1}{2} \bigg[ - 48 -8 +6a \bigg] \\ \\\implies \sf 68 = \frac{1}{2} \bigg[- 56 +6a \bigg] \\ \\\implies \sf 68 = - 28 +3a \\ \\\implies \sf 3a = 68 + 28 \\ \\\implies \sf 3a = 96 \\ \\\implies \sf a = \frac{96}{3} \\ \\ \large\implies\boxed{\boxed{ \sf \blue{ a = 32 \:unit}}}$

Answer 2

Answer:

a = 32

Step-by-step explanation:

given the area of triangle = 68

area of triangle = 1/2 [ X1( Y2 -  Y3 )+ X2 ( Y3- Y1)+ X3(Y1 -Y2) ]

Given points are A( 6,7); B(-4,1); C (a,9)

substituting the no. in the formula

1/2 [ 6 (1-9) + -4 (9-7) + a (7-1) ]  = 68

1/2 [ 6 (-8) -4 (2) + a (6) ] =68

1/2 [ -48 -8 + 6a ]  = 68

1/2  [ 6a - 56 ] = 68

1/2 ×6a-56 = 68

6a - 56 = 68×2

6a - 56 = 136

6a = 136 + 56

6a = 192

a = 192 / 6

a = 32

∴ a=32

  hope it helps ,thank you

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