Question

Prove that the points (3 0) (6 4) and (-1 3) are the vertices of a right angled isosceles triangle

Answer 1

hence prove this your question answer

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption

P(3, 0), Q(6, 4) and R(- 1, 3)

Using distance formula,

PQ = √(x2 - x1)² + (y2 - y1)²

PQ = √(6 - 3)² + (4 - 0)²

PQ = √(3)² + (4)²

PQ = √(9 + 16)

PQ = √25 units

Now,

QR = √(-1 - 6)² + (3 - 4)²

QR = √(-7)² + (-1)²

QR = √(49 + 1)

QR = √50 units

Now,

PC = √(-1 - 3)² + (3 - 0)²

PR = √(-4)² + (3)²

PR = √(16 + 9)

PR = √25 units

Here,

PQ = PR = √25 .

So,

Triangle is an isosceles triangle

So,

In ∆PQR,

Using Pythagoras theorem

QR² = PQ² + PR²

(√50)² = (√25)² + (√25)²

50 = 25 + 25

50 = 50

Here

QR² = PQ² + PR²

Therefore,

Triangle is a right angled triangle.