If the areaof the AABC is 68 sq. units and thevertices are A (6, 7) *(

1.	If the areaof the AABC is 68 sq. units and thevertices are A (6, 7) *(-4,1) and C (a, 9) taken inorder than find the value of a
Answer» 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 $\implies \sf 68 = \frac{1}{2} \bigg[6(1 - 9) - 4(9 - 7) + a(7 - 1) \bigg] \\ \\\implies \sf 68 = \frac{1}{2} \bigg[6( - 8) - 4(2) + a(6) \bigg] \\ \\\implies \sf 68 = \frac{1}{2} \bigg[ - 48 -8 +6a \bigg] \\ \\\implies \sf 68 = \frac{1}{2} \bigg[- 56 +6a \bigg] \\ \\\implies \sf 68 = - 28 +3a \\ \\\implies \sf 3a = 68 + 28 \\ \\\implies \sf 3a = 96 \\ \\\implies \sf a = \frac{96}{3} \\ \\ \large\implies\boxed{\boxed{ \sf \blue{ a = 32 \:unit}}}$

If the areaof the AABC is 68 sq. units and thevertices are A (6, 7) *(-4,1) and C (a, 9) taken inorder than find the value of a

Answer»

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

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