Question

find the area of triangle whose vertices are p(3/1), q(4,2) and r (4,-1/2)​

Answer 1

Step-by-step explanation:

p(3/1),q(4,2),r(4,-1/2)

Area of triangle formula:

$1 \div 2 | x1(y2- y3) + x2(y3 - y1) + x3(y1 - y2)|$

p(3/1,1), q(4,2), r(4,-1/2)

(x1,y2), (x2,y2),(x3,y3)

1/2|1(2-[-1/2])+4(-1/2-1)+4(1-2)|

1/2|1(2+1/2)+4(-1/2-1)+4(-1)|

1/2|1(5/1)+4(-3/2)+4(-1)|

1/2|5/1-12/2_4|

1/2|5/1-6-4|

1/2|5-10|

1/2|-5|

-2.5

Answer 2

Answer:

The area is 1.25 unit²

Step-by-step explanation:

Since the area of a triangle with the vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ is,

$A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Here,

$x_1 = 3, y_1=1, x_2 = 4, y_2=2, x_3 = 4, y_3 = -1/2=-0.5$

Hence, the area of the triangle is,

$A=\frac{1}{2}|3(2+0.5)+4(-0.5-1)+4(1-2)|$

$=\frac{1}{2}|3(2.5)+4(-1.5) + 4(-1)|$

$=\frac{1}{2}|7.5 - 6 - 4|$

$=\frac{1}{2}\times 2.5$

$=1.25\text{ square unit}$

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