Question

Whose
EX÷ find the area of A ABC Whose
very phy are A (3,2), B (11,8), (8,12)​

Answer 1

Question :

Find the area of a ∆ABC whose vertices are A(3,2), B(11,8) and C(8,12)

ANSWER

Given : -

Vertices of ∆ABC A(3,2), B(11,8) and C(8,12)

Required to find : -

• Area of the ∆ABC ?

Formula used : -

To find the area of the ∆ABC whose vertices are $\sf{A(x_1,y_1)}$, $\sf{B(x_2,y_2)}$, and $\sf{C(x_3,y_3)}$ is

$\green{\sf{Area \ of \ \Delta ABC = \dfrac{1}{2} | x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}}$

Solution : -

Given that;

The vertices of the ∆ABC are A(3,2), B(11,8), C(8,12)

Here,

Let's compare the given co-ordinates with their respective general form

A(3,2) $\sf{ \qquad A(x_1,y_1)}$

B(11,8) $\sf{ \qquad B(x_2,y_2)}$

C(8,12) $\sf{ \qquad C(x_3,y_3)}$

Here,

$\sf x_1 = 3 \quad x_2 = 11 \quad x_3 = 8$

$\sf y_1 = 2 \quad y_2 = 8 \quad y_3 = 12$

Using the formula;

$\green{\sf{Area \ of \ \Delta ABC = \dfrac{1}{2} | x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}}$

Area of ∆ABC = (1)/(2)|3(8-12)+11(12-2)+8(2-8)|

Area of ∆ABC = (1)/(2)|3(-4)+11(10)+8(-6)|

Area of ∆ABC = (1)/(2)|-12+110-48|

Area of ∆ABC = (1)/(2)|50|

Area of ∆ABC = (50)/(2)

Area of ∆ABC = 25 square units

Therefore,

Area of the ∆ABC = 25 units²

Knowledge Enhancer !

Q) what is area of collinearity ?

A) If 2 or more points are collinear then the area of the points is founded to be as zero (0)

Q) What is a Mid-Point Formula ?

A) Mid-point formula is useful to find the ? mid-point co-ordinates of a line

The formula is;

$\red{\sf{ M(x,y) = \bigg\lgroup \dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg\rgroup }}$

Q) What is a distance formula ?

A) The distance between 2 points i.e. $\sf{ A(x_1,y_1) \quad B(x_2,y_2)}$ is $\blue{\sf{ \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2}}}$ units

Answer 2

$\huge{\boxed{\fcolorbox{cyan}{pink}{solution}}}$

Let A =

$\implies (x_{1} , \: y_{1}) = (3,2),B \: = ( x_{2} , y_{2}) = (11,8),C = ( x_{3}, y_{3}) = (8,12)$

be the given points

1. Area of a given triangle is given by

Area of triangle =

$\frac{1}{2} { x_{1}( y_{2} - y_{3})} + x_{2} ( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2})$

Area of given triangle

$\frac{1}{2} [3{8 - 12} + 11 \: \: \: {12 - 2} + 8(2 - 8)]$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =$

$\frac{1}{2} (( - 12) + 110 - 48)$

= 50 square unit

More

theorem of midpoint in geometry

the mid point theorem states that the the line segment in a triangle joining the midpoint of two sides of triangle is said to be parallel to its third side.

Midpoint theorem formula

In coordinate geometry, midpoints theorem refers to midpoint of the line segment in define the co-ordinate points of the midpoint of line segment can be found by taking the average of the coordinates of the given endpoints.

The midpoint formula is use to determine the midpoint between the two given points.