Question

find a it the area of triangle ABC is 5. the coordinate A,B,C are (2,3) , (4,5) and (a,3)

Answer 1

Answer:

Given A(−2,1),B(5,4),C(2,−3)

Area of △ABC=

2

1

∣

∣

∣

∣

∣

∣

∣

∣

−2

5

2

1

4

−3

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=

2

1

[∣(−2)(4+3)−(1)(5−2)+(1)(−15−8)∣]=[∣−14−3−23∣]

⇒ Area of ΔABC =

2

1

×(40)=20 square units

Equation of line BC is y−4=

5−2

4+3

(x−5)⇒7x−3y=23

Length of altitude from A is perpendicular distance of A from line BC

Length of altitude =

7

2

+3

2

∣7(−2)−3(1)−23∣

=

58

40

units