Question

A point (a, b) is called lucky if it is on the line ax + by =169. For example, the point (12, 5) is lucky because it is on the line 12x+5y=169. . What is the distance between a lucky point and the origin?

Answer 1

The distance between a lucky point and origin is 13 units

Step-by-step explanation:

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

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

Given,

a point (a, b) is called lucky if it satisfies

ax + by = 169

Coordinates of origin is (0, 0)

The distance between origin and lucky point

$d=\sqrt{(a-0)^2+(b-0)^2}$

$\implies d=\sqrt{a^2+b^2}$

If the lucky point is (12, 5)

The distance is

$d=\sqrt{12^2+5^2}$

$\implies d=\sqrt{144+25}$

$\implies d=\sqrt{169}$

$\implies d=13$

Hope this answer is helpful.

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Answer 2

Given : A point (a, b) is called lucky if it is on the line ax + by =169.

To find : distance between a lucky point and the origin

Solution:

A point (a, b) is called lucky if it is on the line ax + by =169

as point (a , b) lies on line ax + by = 169

hence putting point in equation of line

=> a(a) + b(b)  = 169

=> a² +b ² = 169

=> a² + b² = 13²

Distance of point ( a , b) from origin

= √(a - 0)² +( b - 0)²

= √a² + b²

=  √13²

= 13

Hence distance between a lucky point and the origin  = 13

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