Question

Find the area of a triangle ABC whose vertices are A(4,-6), B(3, 2) and C (5,2)​

Answer 1

Answer:

use distance formula under root (x2-x1)² +(y2-y1)²

Answer 2

Answer:

Given: AD be the median of ΔABC

BD = DC

Sice D is the mid-point of BC

Using the mid point theorem to find the coordinates of D,

D=[

2

x

2

+x

3

,

2

y

2

+y

3

]

D=[

2

3+5

,

2

−2+2

]=[4,0]

Area of ΔABD=

2

1

∣[(x

1

(y

2

−y

4

)+x

2

(y

4

−y

1

)+x

4

(y

1

−y

2

))]∣

=

2

1

∣[(4(−2−0)+3(0−(−6))+4(−6−(−2)))]∣

=

2

1

∣[−8+18−26]∣

=

2

1

∣[−6]∣=3 Sq.units

Area of ΔADC=

2

1

∣[(x

1

(y

4

−y

3

)+x

4

(y

3

−y

1

)+x

3

(y

1

−y

4

))]∣

=

2

1

∣[(4(0−2)+4(2−(−6))+5(−6−0))]∣

=

2

1

∣[−8+32−30]∣

=

2

1

∣[−6]∣=3 Sq.units

Area od ΔABD=ΔADC

Hence proved that the median of triangle AD divides the triangles into equal areas.

Step-by-step explanation:

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