Question

The area of the triangle formed by the points (-1, 6), (-3, -9) and (1, -3) is ____ sq. units

Answer 1

Given :-

Points (-1,6), (-3, -9) and (1, -3)

To Find :-

Area

Solution :-

We know that

$\sf Area_{\triangle}=\dfrac{1}{2}\big| x_1(y_2-y_3) + x_2(y_3-y_1)+x_3(y_1-y_2)\big|$

Here

• x₁ = -1
• x₂ = -3
• x₃ = 1
• y₁ = 6
• y₂ = -9
• y₃ = -3

$\sf Area=\dfrac{1}{2}\big|-1[(-9)-(-3)] + -3(-3 - 6) + 1[6-(-9)]\big|$

$\sf Area=\dfrac{1}{2} \big|-1(-9+3) + -3(-9)+1(6+9)\big|$

$\sf Area=\dfrac{1}{2} \big|-1(-6)+(-27)+1(15)\big|$

$\sf Area=\dfrac{1}{2} \big|6 - 27+15\big|$

$\sf Area=\dfrac{1}{2}\big|-6\big|$

$\sf Area=\dfrac{1}{2}\times 6$

$\sf Area=3\;units^2$

Answer 2

Hello, Buddy!!

Given:-

• ABC is Triangle
• In which A(-1,6), B(-3,-9) & C(1,-3) are points making the ∆le.

To Find:-

• Area of the Traingle.

Required Solution:-

WKT

Area of ∆le = |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| sq.units

→ |-1[-9-(-3)]+(-3)[-3-6]+1[6-(-9)]|

→ |-1[-9+3]+(-3)[-9]+1[6+9]|

→ |-1(-6)+27+1(15)|

→ |6+27+15|

→ |48|

To Remove Modules Multiple With 1/2

→ 1/2(±48)

→ ±24

Area Can't be in negative sence.

✯ Area of Triangle ☞ 24sq.units.

• The area of the triangle formed by the points (-1, 6), (-3, -9) and (1, -3) is 24 sq. units.

@MrMonarque♡

Hope It Helps You ✌️