IDENTIFY THE TYPE OF TRIANGLE FROM THE GIVEN COORDINATES

Problem 1 :

Name the type of triangle formed by the points A(-5, 6), B(-4, -2) and C(7, 5).

Solution:

The three vertices of the triangle are A(-5, 6), B(-4, -2) and (7, 5)

By distance formula,

By distance formula,

By distance formula,

Length of all sides of the triangle are different.

So, ABC is a scalene triangle.

Problem 2 :

Find the coordinates of the point Q on the x axis which lies on the perpendicular bisector of the line segment joining the points A(-5, -2) and B(4, -2). Name the type of triangle formed by the points Q, A and B.

Solution:

Let the points as Q(x, y) = (x, 0) which is equidistant from A(-5, -2) and B(4, -2).

The distance between two points P(x₁, y₁) and Q(x₂, y₂) is 

Take square on both sides,

(-5 - x)² + 4 = (4 - x)² + 4

(-5 - x)² = (4 - x)²

25 + 10x + x² = 16 - 8x + x²

18x + 9 = 0

18x = -9

x = -9/18

x = -1/2

Therefore, the point Q is (-1/2, 0).

AQ = BQ

So, ΔQAB is an isosceles triangle.

Problem 3 :

If the vertices of triangle ABC are A(-2, 4), B(-2, 8) and C(-5, 6) then triangle ABC is classified as 

a) right        b)  scalene      c)  isosceles      d)  equilateral

Solution :

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

A(-2, 4), B(-2, 8)

= √(8 - 4)² + (-2 - (-2))²

= √4² + (-2 + 2)²

= √4² + 0

= 4

B(-2, 8) and C(-5, 6)

= √(6 - 8)² + (-5 - (-2))²

= √(-2)² + (-5 +2)²

= √4 + (-3)²

= √4 + 9

= √13

C(-5, 6) and A(-2, 4)

= √(-2 - (-5))² + (4 - 6)²

= √(-2 + 5)² + (-2)²

= √3² + (-2)²

= √9 + 4

= √13

Since the two sides are equal, it must be a isosceles triangle.

Problem 4 :

Triangle ABC has vertices with A(x, 3), B(−3, −1), and C(−1, −4). Determine and state a value of x that would make triangle ABC a right triangle. Justify why ABC is a right triangle. [The use of the set of axes below is optional.]

Solution :

Slope of AB = (y₂ - y₁) / (x₂ - x₁)

= (-1 - 3) / (-3 - x)

= -4 / (-3 - x)

= 4 / (3 + x)

Slope of BC = (-4 + 1) / (-1 + 3)

= -3 / 2

Slope of CA = (-4 - 3) / (-1 - x)

= -7 / (-1 - x)

= 7 / (1 + x)

Slope of AB (slope of BC) = -1

4 / (3 + x) (-3 / 2) = -1

6/(3 + x) = 1

6 = 3 + x

x = 6 - 3

x = 3

Slope of BC (slope of CA) = -1

(-3/2) (7 / (1 + x)) = -1

21/2(1 + x) = 1

21 = 2(1 + x)

21 = 2x + 2

21 - 2 = 2x

2x = 19

x = 19/2

So, the possible values of x are 3 and 19/2