Question

show that the points (12,8),(-2,6) and (6,0) are the vertices of a right angled triangle and also show that the mid points of the hypotenuse is equidistant from the angular points.

Answer 1

The answer is (16,14).please answer my questions

Answer 2

Let P(12,8), Q(-2,6), and R(6,0) be the given points.

Then,

• By using the Distance formula

PQ² = (x₂ - x₁)² + (y₂ - y₁)²

PQ² = (-2 - 12)² + (6 - 8)²

PQ² = (-14)² + (-2)²

PQ² = 196 + 4

PQ² = 200

PQ = √(200) units

Now,

QR² = (6 + 2)² + (0 - 6)²

QR² = (8)² + (-6)²

QR² = 64 + 36

QR² = 100

QR = √(100)

QR = 10 units

And

RP² = (12 - 6)² + (8 - 0)²

RP² = (6)² + (8)²

RP² = 36 + 64

RP² = 100

RP = √(100)

RP = 10 units

So,

(PQ)² = 200 = 100 + 100 = (QR)² + (RP)²

→ PQ² =  PR² + QR²

Therefore,

• Triangle PQR is a right-angled triangle. The right angle is at angle R.

Now,

• Let S be the midpoint of the hypotenuse PQ.

Then,

The coodinates of S = [ (x₁ + x₂)/2 , (y₁ + y₂)/2 ]

=  [ (12 + (-2))/2 , (8 + 6)/2 ]

= [ (10/2), (14/2) ]

= (5,7)

Now,

SP² = (x₂ - ₁x)² + (y₂ - y₁)²

SP² = (12 - 5)² + (8 - 7)²

SP² = (7)² + (1)²

SP² = 49 + 1

SP² = 50

SP = √50 units

And,

SQ² = (x₂ - ₁x)² + (y₂ - y₁)²

SQ² = (-2 - 5)² + (6 - 7)²

SQ² = (-7)² + (-1)²

SQ² = 49 + 1

SQ² = 50

SQ = √50 units

Also,

SR² = (x₂ - ₁x)² + (y₂ - y₁)²

SR² = (6 - 5)² + (0 - 7)²

SR² = (1)² + (-7)²

SR² = 1 + 49

SR² = 50

SR = √50 units

• Hence, SP = SQ = SR, therefore, 'S' if the midpoints of the hypotenuse. And it is equidistant from the given angular points.