Question

If points P(0, 3) Q(-2, y) and R(-1, 4) are the vertices of a ∆PQR right angled at
P. Find the value of y and also find area of ∆PQR.​

Answer 1

Answer: y = 3  , Area of triangle = 1

Step-by-step explanation:

Let the points are P(0,3) , Q(-2,y) and R(-1,4)

By distance formula PQ = $\sqrt{(y-3)^{2} +(-2-0)^{2} }$

                                       = $\sqrt{(y-3)^{2}+4 }$

By distance formula QR = $\sqrt{(-2+1)^{2}+(y-4)^{2}$

                                      = $\sqrt{(y-4)^{2}+1 }$

By Distance formula PR = $\sqrt{(-1-0)^{2}+(4-3)^{2} }$

                                      = $\sqrt{2}$

By Pythagoras theorem

PR² +QR² = PQ²

2+(y-4)²+1 = (y-3)² +4

2+1 + y² + 16 -8y = y²+9 -6y +4

-2y = -6

y =3

Hypotenuse = PQ = 2

Base = QR = $\sqrt{2}$

Height = PR = $\sqrt{2}$

Area of triangle = 1/2 x b x h

                          = 1/2 x $\sqrt{2}$ x $\sqrt{2}$

                           = 1

Answer 2

Answer:

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