Verify the following:(0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.

Asked by Abhisek | 1 year ago |  115

##### Solution :-

(0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.

Let the points be

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

Firstly let us calculate the distance of PQ, OR and PR

Calculating PQ

P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)

By using the formula,

Distance PQ =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = 0, y1 = 7, z1 = 10

x2 = – 1, y2 = 6, z2 = 6

Distance PQ =

$$\sqrt{(-1 – 0)^2 + (6 – 7)^2 + (6 – 10)^2}$$

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

$$\sqrt{1 + 1 + 16}$$

$$\sqrt{18}$$

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

By using the formula,

Distance QR =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = 1, y1 = 6, z1 = – 6

x2 = 4, y2 = 9, z2 = – 6

Distance QR =

$$\sqrt{(4 – 1)^2 + (9 – 6)^2 + (-6 – (-6))^2}$$

$$\sqrt{(3)^2 + (3)^2 + (-6+6)^2}$$

$$\sqrt{9 + 9 + 0}$$

$$\sqrt{18}$$

Calculating PR

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

By using the formula,

Distance PR =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

So here,

x1 = 0, y1 = 7, z1 = 10

x2 = – 4, y2 = 9, z2 = 6

Distance PR =

$$\sqrt{(-4 – 0)^2 + (9 – 7)^2 + (6 – 10)^2}$$

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

$$\sqrt{16 + 4 + 16}$$

$$\sqrt{36}$$

Now,

PQ2 + QR2 = 18 + 18

= 36

= PR2

By using converse of Pythagoras theorem,

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

Answered by Pragya Singh | 1 year ago

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