Question

find the area,if the triangle whose vertices are the points (3,4),(5,7) and (-2,-3)​

Answer 1

Answer:

7

Area of a triangle whose vertices are (X1,Y1) , (X2,Y2) and (X3,Y3) is

= 1/2×|X1(Y2-Y3)+X2(Y3-Y1)+X3(Y1-Y2)|

= 1/2×|3(7-(-3))+5(-3-7)+(-2)(4-7)|

= 1/2×|3(7+3)+(5×-10)-(2×-3)|

= 1/2×|30-50+6|

= 1/2×|-14|

= 1/2×14

= 7

Answer 2

The area of triangle ,if the triangle whose vertices are the points (3,4),(5,7) and (-2,-3)​ is 0.5 square units.

Step-by-step explanation:

Given:

The triangle whose vertices are the points (3,4),(5,7) and (-2,-3)​.

To Find:

The area,if the triangle whose vertices are the points (3,4),(5,7) and (-2,-3).

Formula Used:

If vertices  of a  triangle are the points (a,b),(c,d) and (e,f)​.

Area of triangle $=\frac{1}{2} [a(d-f)+c(f-b)+e(b-d))$   ---- formula no.01.

Solution:

As given,the triangle whose vertices are the points (3,4),(5,7) and (-2,-3)​.

Let A point vertices (a,b)=(3,4)

Let B point vertices (c,d)=(5,7)

Let C point vertices (a,b)=(-2,-3)

Applying formula no.01.

Area of triangle $=\frac{1}{2} [3(7-(-3))+5((-3)-4)+(-2)(4-7)]$

                          $=\frac{1}{2} [3(7+3)+5(-3-4)-2(4-7)]$

                         $=\frac{1}{2} [3(10)+5(-7)-2(-3)]$

                         $=\frac{1}{2} [30-35+6]$

                         $=\frac{1}{2} [1]\\=0.50$ square units.

Thus,the area of triangle ,if the triangle whose vertices are the points (3,4),(5,7) and (-2,-3)​ is 0.5 square units.

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