Q. 3 B4.5( 41 Votes )

# Verify the follow

Answer :

Let points be

P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)

__Calculating PQ__

P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)

Distance PQ

Here,

x_{1} = 0, y_{1} = 7, z_{1} = 10

x_{2} = – 1, y_{2} = 6, z_{2} = 6

Distance PQ

__Calculating QR__

Q ≡ (– 1, 6, 6) and R ≡ (– 4, 9, 6)

Distance QR

Here,

x_{1} = – 1, y_{1} = 6, z_{1} = 6

x_{2} = – 4, y_{2} = 9, z_{2} = 6

Distance QR

__Calculating PR__

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

Distance PR

Here,

x_{1} = 0, y_{1} = 7, z_{1} = 10

x_{2} = – 4, y_{2} = 9, z_{2} = 6

Distance PR

Now,

PQ^{2} + QR^{2} = 18 + 18 = 36 = PR^{2}

Hence, According to converse of pythagoras theorem,

the given vertices P, Q & R are the vertices of a right – angled triangle at Q.

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