# AREA OF TRIANGLE WITH 3 COORDINATES

Problem 1 :

Find the area of a triangle whose vertices are (3, 0), (7, 0) and (8, 4).

Solution:

The vertices of the given triangle A(3, 0), B(7, 0) and C(8, 4).

x1 = 3, y1 = 0, x2 = 7, y2 = 0, x3 = 8, y3 = 4

Area of a  triangle

Area can't be negative.

Therefore, the area of the triangle is 20 square units.

Problem 2 :

The area of a triangle whose vertices are (5, 0), (8, 0) and (8, 4) (in sq.units) is

A) 20    B) 12    C) 6

Solution:

The vertices of the given triangle A(5, 0), B(8, 0) and C(8, 4)

x1 = 5, y1 = 0, x2 = 8, y2 = 0, x3 = 8, y3 = 4

Area of a  triangle

So, option (C) is correct.

Problem 3 :

The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is (7/2, y), find the value of y.

Solution:

Let A(2, 1), B(3, -2) and C(7/2, y)

x1 = 2, y1 = 1, x2 = 3, y2 = -2, x3 = 7/2, y3 = y

Area of a triangle

Problem 4 :

Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.

Solution:

The vertices of the given triangle A(1, -1), B(-4, 2k) and C(-k, -5)

x1 = 1, y1 = -1, x2 = -4, y2 = 2k, x3 = -k, y3 = -5

Area of a  triangle

Problem 5 :

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Solution:

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).

Hence, the area of the required triangle is 1 sq. unit.

Problem 6 :

For what type of k, (k > 0), is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) to 53 sq. units?

Solution:

The vertices of the given triangle A(-2, 5), B(k, -4) and C(2k + 1, 10)

x1 = -2, y1 = 5, x2 = k, y2 = -4, x3 = 2k + 1, y3 = 10

Area of a  triangle

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