Question

area of the triangle ABC with coordinates of A as( 1 ,- 4 )and the coordinates of the midpoints of sides AB and AC respectively are (2, -1) and (0 ,- 1)​

Answer 1

Answer:

Area = 7.021

Step-by-step explanation:

Coordinate of point A on triangle ABC is (1, -4)

Mid point of AB is (2, -1)

Mid point of AC is (0, -1)

Let (x,y) is point B.  

Hence distance between (1, -4), (2, -1)  and (2, -1) and (x , y) is same.

(2 -1)² + (-1 + 4)² = (x – 2)² + (y + 1)²

X² – 4x + 4 + y² +2y + 1 = 10

X² – 4x  + y² +2y = 5  --------------------------------E1

Slopes between (1, -4), (2, -1)  and (2, -1) and (x , y) are same

3/2 = (y+1)/(x -2)

2y + 2 = 3x – 4

3x – 2y = 6

Substituting above equation in E1 gives us,  

X² – 4x + (3x – 6)²/4 + (3x – 6) = 5

X² – 4x + (9x² -36x + 36)/4 + (3x – 6) = 5

13x² -42x + 7 = 0

x=3.05448 or 0.176286

y = 1. 577 or -2.7357

B is (3.05, 1.577)

Let (x ,y) is point c.

Hence distance between (1, -4), (0, -1)  and (0, -1) and (x , y) is same.

(0 -1)² + (-1 + 4)² = (x – 0)² + (y + 1)²

X² + y² +2y + 1 = 10

X²  + y² +2y = 9  --------------------------------E2

Slopes between (1, -4), (0, -1)  and (0, -1) and (x , y) are same

3/-1 = (y+1)/(x -0)

-y - 1 = 3x  

3x + y = -1

Substituting above equation in E2 gives us,  

X² + (-3x – 1)² + (-3x – 1) = 9

10x² +3x -9 = 0

X = 0.810469 or -1.11047

Y = -3.43 or -2.89

Pont is C is (-1.11, -2.89)

Area of a triangle with vertices is given by  

1/ 2 [x1(y2–y3) + x2 (y3–y1) + x3 (y1–y2)]

Substituting we get

Area = 7.021

Answer 2

Answer:

12

Step-by-step explanation:

area of the triangle ABC with coordinates of A as( 1 ,- 4 )and the coordinates of the midpoints of sides AB and AC respectively are (2, -1) and (0 ,- 1)​

A = (1 , -4)

B = (Bx , By)

C = (cx , Cy)

midpoint of AB = (Bx+1)/2  , (By - 4)/2   = (2, -1)  given

Bx = 3  & By = 2

midpoint of AC = (Cx+1)/2  , (Cy - 4)/2   = (0 , -1)  given

Cx = -1  & Cy = 2

AB² =  (3-1)² + (2-(-4))² = 4 + 36 = 40

BC² = (-1-3)² + (2-2)² = 16 + 0 = 16

AC² = (-1-1)² + (2-(-4))² = 4 + 36 = 40

AB = 2√10  AC = 2√10 BC = 4

s = (AB + AC + BC)/2

s = 2√10 + 2 = 2(√10 + 1)

Area = √(s)(s-AB)(s-BC)(s-AC)

= √(2(√10 + 1)(2)(2)2(√10 - 1) )

= 4 √(10 - 1)

= 4 √9

= 4 ×3

= 12