Question

What is distance between line 12x+35y+5=0 and point (2,0).

Answer 1

29 / 37

Explanation:

The distance between a point (p, q) and a line ax + by + c = 0 is:

d = | ap + bq + c | / √(a² + b²)

Here, a = 12, b = 35, c = 5, p = 2, and q = 0.

d = | (12)(2) + (35)(0) + 5 | / √(12² + 35²)

d = 29 / 37

Answer 2

The distance between line 12x+35y+5=0 and point (2,0) is
`(29)/(37)` units.

Explanation:

The distance between a line
`Ax+By+C=0` and a point
`(x_0,y_0)` is

`d=(Ax_0+By_0+C)/(√(A^2+B^2))`

The distance between line 12x+35y+5=0 and point (2,0) is

`d=(12(2)+35(0)+5)/(√((12)^2+(35)^2))`

`d=(24+0+5)/(√(144+1225))`

On further simplification we get

`d=(24+0+5)/(√(1369))`

`d=(29)/(37)`

Therefore, the distance between line 12x+35y+5=0 and point (2,0) is
`(29)/(37)` units.