Question

if A (-5,7) B (-4,-5) C(4,5) are the vertices of triangle.Find the area of triangle ABC.

Answer 1

Answer:

Area of triangle ABC is 53 sq units

Step-by-step explanation:

Since we know,

Area of triangle having 3 vertices is :

( x₁ , y₁ ) , ( x₂ , y₂ ) and ( x₃ , y₃ ) is given by formula :

Area = $\frac{1}{2} [ x_{1} (y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Given,

( $x_{1} , y_{1}$ ) = ( -5 , 7 )

( $x_{2} , y_{2}$ ) = ( -4 , -5 )

( $x_{3} , y_{3}$ ) = ( 4 , 5 )

Substituting the values we get,

=  $\frac{1}{2} [ -5 ((-5)-5)+(-4)(5-7)+4(7-(-5))]$

=  $\frac{1}{2} [ -5 (-10)+(-4)(-2)+4(12)]$

=  $\frac{1}{2} [ 50+8+48]$

= $\frac{1}{2} [ 106]$

= 53

Hence,

Area of triangle ABC is 53 sq units

Answer 2

Step-by-step explanation:

Since we know,

Area of triangle having 3 vertices is :

( x₁ , y₁ ) , ( x₂ , y₂ ) and ( x₃ , y₃ ) is given by formula :

Area = \frac{1}{2} [ x_{1} (y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

Given,

( x_{1} , y_{1}x

1

,y

1

) = ( -5 , 7 )

( x_{2} , y_{2}x

2

,y

2

) = ( -4 , -5 )

( x_{3} , y_{3}x

3

,y

3

) = ( 4 , 5 )

Substituting the values we get,

= \frac{1}{2} [ -5 ((-5)-5)+(-4)(5-7)+4(7-(-5))]

2

1

[−5((−5)−5)+(−4)(5−7)+4(7−(−5))]

= \frac{1}{2} [ -5 (-10)+(-4)(-2)+4(12)]

2

1

[−5(−10)+(−4)(−2)+4(12)]

= \frac{1}{2} [ 50+8+48]

2

1

[50+8+48]

= \frac{1}{2} [ 106]

2

1

[106]

= 53

Hence,

Area of triangle ABC is 53 sq units.

hope it will help you