Show that the points (-2,3,5),(1,2,3) and (7,0,−1) are collinear.

Asked by Abhisek | 1 year ago |  98

##### Solution :-

If three points are collinear, then they lie on a line.

Firstly let us calculate distance between the 3 points

i.e. PQ, QR and PR

Calculating PQ

P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)

So here,

x1 = – 2, y1 = 3, z1 = 5

x2 = 1, y2 = 2, z2 = 3

Distance PQ =

$$\sqrt{(1 – (-2))^2 + (2 – 3)^2 + (3 – 5)^2}$$

$$\sqrt{(3)^2 + (-1)^2 + (-2)^2}$$

$$\sqrt{9 + 1 + 4}$$

$$\sqrt{14}$$

Calculating QR

Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

So here,

x1 = 1, y1 = 2, z1 = 3

x2 = 7, y2 = 0, z2 = – 1

Distance QR =

$$\sqrt{(7 – 1)^2 + (0 – 2)^2 + (-1 – 3)^2}$$

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

$$\sqrt{36 + 4 + 16}$$

= $$\sqrt{56}$$

= $$2\sqrt{14}$$

Calculating PR

P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)

So here,

x1 = – 2, y1 = 3, z1 = 5

x2 = 7, y2 = 0, z2 = – 1

Distance PR =

$$\sqrt{(7 – (-2))^2 + (0 – 3)^2 + (-1 – 5)^2}$$

$$\sqrt{(9)^2 + (-3)^2 + (-6)^2}$$

$$\sqrt{81 + 9 + 36}$$

$$\sqrt{126}$$

$$3\sqrt{14}$$

Thus, PQ = $$\sqrt{14}$$, QR = $$2\sqrt{14}$$ and PR = $$3\sqrt{14}$$

So, PQ + QR =$$\sqrt{14}+ 2\sqrt{14}$$

$$3\sqrt{14}$$

= PR

The points P, Q and R are collinear.

Answered by Pragya Singh | 1 year ago

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