Question

Area of the triangle formed by $\left(x_{1,}{y}_1\right),(x_2,y_2),(3y_2,(-2y_1\left)\right)$

• A. $\frac{1}{2}[x_1{y}_2+2x_1{y}_1-3x_2{y}_1+3y_2{y}_1-3{y_2}^2]$
• B. $\frac{1}{2}[x_1{y}_2+2x_1{y}_1-3x_2{y}_1+3y_2{y}_1-3{y_2}^3]$
• C. $\frac{2}{3}[x_1{y}_2+2x_1{y}_1-3x_2{y}_1+3y_2{y}_1-3{y_2}^2]$
• D. $=\frac{1}{2}[x_1{y}_2+2x_1{y}_1-3x_2{y}_1+3y_2{y}_3]$
  

Answer 1

Answer: (A)

$\frac{1}{2}[x_1{y}_2+2x_1{y}_1-3x_2{y}_1+3y_2{y}_1-3{y_2}^2]$

We have,

$A(x_1,y_1)=(x_1,y_1)$

$B(x_2,y_2)=(x_2,y_2)$

$C(x_3,y_3)=(3y_2,-2y_1)$

We know that the area of triangle is

$Areaof\Delta ABC=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\frac{1}{2}[x_1(y_2+2y_1)+x_2(-2y_1-y_1)+3y_2(y_1-y_2)]$

$=\frac{1}{2}[x_1{y}_2+2x_1{y}_1-3x_2{y}_1+3y_2{y}_1-3{y_2}^2]$

Hence, the is the answer