Question

9. Find the area of the triangle ABC whose vertices are A(2,3); B (-1,0) and
C (2,-4).​

Answer 1

Explanation:

To find : The area of triangle ABC whose vertices are A(2,3) B(-1,0)and C(2,-4) ? Therefore, the area of the triangle ABC is 10.5 square unit

Answer 2

Answer:

Hence, the area of given triangle = 21/2 units = 10.5 units

Explanation:

Given, that,

Coordinates : A(2,3) , B(-1,0) and C(2,-4)

So here,

x1=2    I     x2=-1     I   x3=2

I I

y1=3   I     y2=0    I    y3= -4

We know that ,

For coordinate geometry,

Area of Traingle = 1/2 ×  {x1(y2-y3) + x2(y3-y1) + x3(y1-y2)}

So By applying the values in the formula, we get,

Area of Triangle = 1/2 {2(0-(-4)) + (-1) (-4-3) + 2(3-0)}

=> Area of Triangle = 1/2 {8-(-7)+6}

=> Area of Triangle = 1/2 {8+7+6}

=> Area of Triangle = 1/2 × 21

=> Area of Triangle = 21/2

Hence, the area of given triangle = 21/2 units = 10.5 units

"A piece of Additional Information " :

In coordinate geometry for 2 dimensional figures, we have,

Four more formulas :

• To find the distance between the points AB

=> AB = √(x2 - x1)^2 + (y2 - y1)^2

• To find the coordinates point intersecting the line segment AB in the ratio of m : n.

=> x = (mx2 + nx1) / (m + n)

=> y = (my2 + ny1) / (m + n)

• To find the coordinates of the point bisecting the line segment AB :

=> x = (x1 + x2) / 2

=> y = (y1 + y2) / 2

• For the coordinate of centroid of triangle ABC :

=> x = (x1 + x2 + x3) / 3

=> y = (y1 + y2 + y3) / 3