Question

find the area of the triangle whose vertices are (1,-1),(-4,6)and (-3,-5)

Answer 1

Answer:

Area of the triangle formed by given 3 points is 24 square units

Step-by-step explanation:

Formula used:

Area of the triangle formed by the points

$(x_1, y_1),\; (x_2, y_2) \:and \:(x_3, y_3)\:is\\\\|\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\:\:square\:units$

Given points are (1, -1), (-4, 6)and (-3, -5)

Area of the triangle

$=|\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\:\:square\:units\\\\=|\frac{1}{2}[1(6+5)-4(-5+1)-3(-1-6)|\\\\=|\frac{1}{2}[1(11)-4(-4)-3(-7)|\\\\=|\frac{1}{2}[11+16+21]|\\\\=|\frac{1}{2}[48]|\\\\=24\:square\:units$

Answer 2

Answer:

24

Step-by-step explanation:

find the area of the triangle whose vertices are (1,-1),(-4,6)and (-3,-5)

First Find length of 3 Sides

Side1² =  (-4-1)² + (6-(-1))² = (-5)² + 7² = 25 + 49 = 74

Side2² = (-3-1)² + (-5 -(-1))² =  (-4)² + (-4)² = 16 + 16 = 32

Side3² = (-3 -(-4))² + (-5 - 6)² = 1² + (-11)² = 1 + 121 = 122

Side 1 = √74 = 8.60

Side 2 = √32 = 5.66

Side 3 = √122 = 11.05

S = (side 1 + Side2 + side3)/2 = 25.31/2 = 12.66

Area = √(s(s-side1)(s-side2)(s-side3)

= √(12.66 * 4.06 * 7 * 1.61)

= √579.27

= 24