Question

find the area of triangle ABC whose coordinates are A(4,-6)B(3,-2)C(5,2) then find the length of the median AD​

Answer 1

✪ Given :-

• Coordinates of A = ( 4 , -6 )
• Coordinates of B = ( 3 , -2 )
• Coordinates of C = ( 5 , 2 )

✪ To Find :-

• Area of the triangle ABC

✪ Solution :-

$\red\bigstar\: \boxed{\bf \green{Area = \frac{1}{2} \bigg[x_1(y_2 - y_3) + x_2(y_3 - y_1) +x_3(y_1 - y_2) \bigg ]}}$

Here

• x₁ = 4
• x₂ = 3
• x₃ = 5
• y₁ = -6
• y₂ = -2
• y₃ = 2

Substitute values in formula

$\longmapsto \sf Area = \frac{1}{2} \bigg[4(-2-2) + 3(2 + 6) +5( - 6 + 2) \bigg ]$

$\longmapsto \sf Area = \frac{1}{2} \bigg[4(-4) + 3(8) +5( - 4) \bigg ]$

$\longmapsto \sf Area = \frac{1}{2} \bigg[ - 16+24 - 20\bigg ]$

$\longmapsto \sf Area = \frac{1}{2} \bigg[ - 12\bigg ]$

$\longmapsto\boxed{\sf\purple{ Area = - 6 \: {unit}^{2}} }$

Answer 2

Answer:

AD is the median of ABC from vertex A.

Since area can not the negative Area of triangle.

ACB = 3 square units.

From (i) and (ii) Area triangle ADB = Area triangle ACB it is verified that median of triangle ABC divides it into two triangle of equal areas.

Step-by-step explanation:

Plz mark as Brainliest...