Question

Find the distance between the points (1, 2), and (4, 6).

Answer 1

Gívєn :

Find the distance between the points (1,2) and (4,6)

fσrmulαs usєd :

Distance between two points :

The distance between any two points in the plane is the length of the line segment joining them.

The distance between two points P $\sf\:x_{1},y_{1}$ and Q $\sf\:x_{2},y_{2}$ is given by

$\sf\:PQ=\sqrt{(x_{2}-x_{1}){}^{2}+(y_{2}-y_{1}){}^{2}}$

Sσlutíσn :

Let A (1,2) and B(4,6)

Distance between points A and B

= AB

$\sf = \sqrt{(4 - 1) {}^{2} + (6 - 2) {}^{2} }$

$\sf = \sqrt{3 {}^{2} + 4 {}^{2} }$

$\sf = \sqrt{9 + 16}$

$\sf = \sqrt{25} = 5 \: units$

Therefore,the distance between points (1,2) and (4,6) is 5 units .

Answer 2

Given :  points ( 1 , 2 ) and ( 4 , 6 )

To Find :  the distance between points ( 1 , 2 ) and ( 4 , 6 )

Solution:

Distance between point (x₁ , y₁ )  and (x₂ , y₂)  is given by

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

points ( 1 , 2 ) and ( 4 , 6 )

x₁ = 1

x₂ = 4

y₁ = 2

y₂ = 6

the distance between points ( 1 , 2 ) and ( 4 , 6 )

 $\sqrt{(4-1)^2+(6-2)^2}$

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

$=\sqrt{9+16}$

$=\sqrt{25}$

= ±5

as Distance can not be negative

Hence the distance between points ( 1 , 2 ) and ( 4 , 6 )  is 5

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