VERIFYING GEOMETRIC PROPERTIES OF TRIANGLE WITH GIVEN POINTS

Problem 1 :

Is ∆PQR a right angled triangle if P(1, 2), Q(5, -1) and R(-4, -4) ?

Solution :

P(1, 2), Q(5, -1) and R(-4, -4) 

D = √[(x₂ - x₁)² + (y₂ - y₁)²]

 P(1, 2) and Q = (5, -1)

x₁ = 1, y₁ = 2, x₂ = 5, y₂ = -1

PQ = √[(5 - 1)² + (-1 - 2)²]

= √[(4)² + (-3)²]

= √[16 + 9]

= √25

PQ = 5

 Q = (5, -1) and  R(-4, -4) 

x₁ = 5, y₁ = -1, x₂ = -4, y₂ = -4

QR = √[(-4 - 5)² + (-4 + 1)²]

= √[(-9)² + (-3)²]

= √[81 + 9]

QR = √90

R(-4, -4)  and  P(1, 2)

x₁ = -4, y₁ = -4, x₂ = 1, y₂ = 2

RP = √[(1 + 4)² + (2 + 4)²]

= √[(5)² + (6)²]

= √[25 + 36]

RP = √61

∆PQR is not a right angled triangle.

Problem 2 :

Classify  ∆DEF following triangle as scalene, isosceles or equilateral if D(5, 2), E(-3, 4) and F(-2, -3).

Solution :

D(5, 2), E(-3, 4) and F(-2, -3).

D = √[(x₂ - x₁)² + (y₂ - y₁)²]

D(5, 2) and E(-3, 4)

x₁ = 5, y₁ = 2, x₂ = -3, y₂ = 4

DE = √[(-3 - 5)² + (4 - 2)²]

= √[(-8)² + (2)²]

= √[64 + 4]

DE = √68

E(-3, 4) and F(-2, -3).

x₁ = -3, y₁ = 4, x₂ = -2, y₂ = -3

EF = √[(-2 + 3)² + (-3 - 4)²]

= √[(1)² + (-7)²]

= √[1 + 49]

EF = √50

F(-2, -3) and D(5, 2)

x₁ = -2, y₁ = -3, x₂ = 5, y₂ = 2

FD = √[(5 + 2)² + (2 + 3)²]

= √[(7)² + (5)²]

= √[49 + 25]

FD = √74

Since all the sides are having different length, it must be a scalene triangle.

Problem 3 :

Given the points X(1, 4), Y(-2, 2) and Z(3, 1). Verify that ∆XYZ is a right triangle.

Solution :

X(1, 4), Y(-2, 2) and Z(3, 1)

D = √[(x₂ - x₁)² + (y₂ - y₁)²]

X(1, 4) and Y(-2, 2)

x₁ = 1, y₁ = 4, x₂ = -2, y₂ = 2

XY = √[(-2 - 1)² + (2 - 4)²]

= √[(-3)² + (-2)²] 

= √[9 + 4]

XY = √13 

Y(-2, 2) and Z(3, 1)

x₁ = -2, y₁ = 2, x₂ = 3, y₂ = 1

YZ = √[(3 + 2)² + (1 - 2)²]

= √[(5)² + (-1)²] 

= √[25 + 1]

YZ = √26 

Z(3, 1) and X(1, 4)

x₁ = 3, y₁ = 1, x₂ = 1, y₂ = 4

ZX = √[(1 - 3)² + (4 - 1)²]

= √[(-2)² + (3)²] 

= √[4 + 9]

ZX = √13

Verifying Pythagorean theorem :

(YZ)² = (XY)² + (ZX)²

(√26)² = (√13)² + (√13)²

26 = 13 + 13

26 = 26

Since the lengths of triangle satisfies Pythagorean theorem, it must be right triangle.

∴ ∆XYZ is not a right triangle. It is a isosceles triangle.

Problem 4 :

A triangle has vertices K(-2, 2), L(1, 5) and M(3, -3). Verify that :

a) The triangle has a right angle.

b) The midpoint of the hypotenuse is the same distance from each vertex.

Solution :

a)

Given, K(-2, 2), L(1, 5) and M(3, -3)

D = √[(x₂ - x₁)² + (y₂ - y₁)²]

K(-2, 2) and L(1, 5)

x₁ = -2, y₁ = 2, x₂ = 1, y₂ = 5 

KL = √[(1 + 2)² + (5 + 2)²]

= √[(3)² + (7)²]

= √[9 + 49]

KL = √58

L(1, 5) and M(3, -3)

D = √[(x₂ - x₁)² + (y₂ - y₁)²]

x₁ = 1, y₁ = 5, x₂ = 3, y₂ = -3 

LM = √[(3 - 1)² + (-3 - 5)²]

= √[(2)² + (-8)²]

= √[4 + 64]

LM = √68

M(3, -3) and K(-2, 2)

x₁ = 3, y₁ = -3, x₂ = -2, y₂ = 2 

MK = √[(-2 - 3)² + (2 + 3)²]

= √[(-5)² + (5)²]

= √[25 + 25]

LM = √50

b) L(1, 5) and M(3, -3)

Suppose that the midpoint is P.

x₁ = 1, y₁ = 5, x₂ = 3, y₂ = -3 

P = (2, 1)

Then,  K(-2, 2) and P (2, 1)

x₁ = -2, y₁ = 2, x₂ = 2, y₂ = 1 

KP = √[(2 + 2)² + (1 - 2)²]

= √[(4)² + (-1)²]

= √[16 + 1]

KP = √17

L(1, 5) and P (2, 1)

x₁ = 1, y₁ = 5, x₂ = 2, y₂ = 1 

LP = √[(2 - 1)² + (1 - 5)²]

= √[(1)² + (-4)²]

= √[1 + 16]

LP = √17

M(3, -3) and P (2, 1)

x₁ = 3, y₁ = -3, x₂ = 2, y₂ = 1 

MP = √[(2 - 3)² + (1 + 3)²]

= √[(-1)² + (4)²]

= √[1 + 16]

MP = √17

So, the midpoint of the hypotenuse is the same distance from each vertex.

Problem 5 :

A triangle has vertices U(5, 5), V(1, -3) and W(-3, -1). Verify that :

a) ∆UVW is a right triangle

b) The median from the right angle to the hypotenuse is half  as long as the hypotenuse.

Solution :

a)

Given, U(5, 5), V(1, -3) and W(-3, -1)

D = √[(x₂ - x₁)² + (y₂ - y₁)²]

U(5, 5), V(1, -3)

x₁ = 5, y₁ = 5, x₂ = 1, y₂ = -3 

UV = √[(1 - 5)² + (-3 - 5)²]

= √[(-4)² + (-8)²]

= √[16 + 64]

= √80

V(1, -3) and W(-3, -1)

x₁ = 1, y₁ = -3, x₂ = -3, y₂ = -1 

VW = √[(-3 - 1)² + (-1 + 3)²]

= √[(-4)² + (2)²]

= √[16 + 4]

= √20

W(-3, -1) and U(5, 5)

x₁ = -3, y₁ = -1, x₂ = 5, y₂ = 5 

WU = √[(5 + 3)² + (5 + 1)²]

= √[(8)² + (6)²]

= √[64 + 36]

WU = √100

Verifying Pythagorean theorem :

(WU)² = (VW)² + (UV)²

(√100)² = (√20)² + (√80)²

100 = 20 + 80

100 = 100

Since it satisfies the Pythagorean theorem, it must create a right triangle.

b) From the above proof, we understand that UW is the hypotenuse.

U(5, 5) and W(-3, -1)

Midpoint of UW = (5 - 3)/2, (5 - 1)/2

= 2/2, 4/2

Midpoint of hypotenuse = (1, 2)

Distance between V and the midpoint  :

V(1, -3) and midpoint (1, 2)

Distance between them = √[(1-1)² + (2+3)²]

= √[0 + 5²

= 5

Length of hypotenuse = 10

Length of median = 5

Hence it is proved.