Question

find the area of triangle whose the ventic of A (1,-1 ) B ( -4,6 ) C ( -3,5)​

Answer 1

Given Vertices

A(1,-1) , B(-4,6) and C(-3 , 5)

To Find

Area of ΔABC

Formula

ΔABC =| 1/2[x₁(y₂ - y₃)+x₂(y₃ - y₁) + x₃(y₁ - y₂)]|

Now we have

x₁ = 1 , y₁= -1 , x₂ = -4 , y₂ = 6 ,x₃ = -3 and y₃ = 5

Put the value on formula

ΔABC =| 1/2[1(6-5)-4(5+1)-3(-1-6)]|

ΔABC =| 1/2[1(1) - 4(6) -3(-7)]|

ΔABC = |1/2[1-24+21]|

ΔABC = |1/2[22-24]|

ΔABC = |1/2[-2]|

ΔABC = |-1|

ΔABC = = 1 Sq units

Answer

Area of  ΔABC is 1sq units

Answer 2

${\large{\pmb{\sf{\underline{Explanation...}}}}}$

★ We have to find out the area of triangle whose vertices are given as A(1,-1); B(-4,6) and C(-3,5).

${\large{\pmb{\sf{\underline{Using \: formulas...}}}}}$

★ Formula to find out the area of triangle whose vertices are given =

${\small{\underline{\boxed{\sf{\dfrac{1}{2} [x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]}}}}}$

${\large{\pmb{\sf{\underline{Where...}}}}}$

★ x₁ and y₁ are the 1st vertice

★ x₂ and y₂ are the second vertice

★ x₃ and y₃ are the third vertice

${\large{\pmb{\sf{\underline{Here...}}}}}$

★ x₁ is 1 here

★ x₂ is -4 here

★ x₃ is -3 here

★ y₁ is -1 here

★ y₂ is 6 here

★ y₃ is 5 here

${\large{\pmb{\sf{\underline{Full \; Solution...}}}}}$

${\small{\underline{\boxed{\sf{\bull \: \: \dfrac{1}{2} [x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]}}}}} \\ \\ :\implies \sf \dfrac{1}{2} [x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})] \\ \\ :\implies \sf \dfrac{1}{2} [1(6-5) + -4(5-(-1)) + -3(-1-6)] \\ \\ :\implies \sf \dfrac{1}{2} [1(1) + -4(6) -3(-7)] \\ \\ :\implies \sf \dfrac{1}{2} [1(1) + (-24) +21)] \\ \\ :\implies \sf \dfrac{1}{2} [1 - 24 + 21] \\ \\ :\implies \sf \dfrac{1}{2} [1 - 3] \\ \\ :\implies \sf \dfrac{1}{2} (-2) \\ \\ :\implies \sf -1 \\ \\ :\implies \sf Henceforth, \: 1 \: unit \: sq. \: is \: area \: of \: \triangle$

${\large{\pmb{\sf{\underline{Additional \; Knowledge...}}}}}$

$\underline{\bigstar\:\textsf{Distance Formula\; :}}$

Distance formula is used to find the distance between two given points.

${\underline{\boxed{\sf{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$ ⠀

$\underline{\bigstar\:\textsf{Section Formula\; :}}$

Section Formula is used to find the co ordinates of the point(Q) Which divides the line segment joining the points (B) and (C) internally or externally.

${\underline{\boxed{\sf{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$ ⠀

$\underline{\bigstar\:\textsf{Mid Point Formula\; :}}$

Mid Point formula is used to find the mid points on any line.

${\underline{\boxed{\sf{\quad \Bigg(\dfrac{x_1 + x_2}{2} \; or\; \dfrac{y_1 + y_2}{2} \Bigg)\quad}}}}$