Area of Triangle Problem | AMC-8, 2019 | Problem 21

Find the three vertex of the triangle

Can you now finish the problem ..........

The area of the Triangle =\(\frac{1}{2} \times \{x_1(y_2 - y_3)+x_2(y+3 -y_1)+x_3(y_1 -y_2)\}\)

can you finish the problem........

Solving two The lines y=5 and y=1+x are intersect at (4,5)=\((x_1,y_1)\)(say)

Solving two The lines y=5 and y=1-x are intersect at (-4,5)=\((x_2,y_2)\)(say)

Solving two The lines y=1-x and y=1+x are intersect at (0,1)=\((x_1,y_1)\)(say)

Then the area of Triangle =\(\frac{1}{2} \times\{ x_1(y_2 - y_3)+x_2(y+3 -y_1)+x_3(y_1 -y_2)\}\)

= \(\frac{1}{2} \times \{4(5-1)+(-4)(1-5)+0(5-5)=\frac{1}{2} (16+16)=16\}\)