Question

What is the shortest distance between the line given by -2x+3y+4=0 and the point (5,6)?

Answer 1

Answer:

12/√13

Step-by-step explanation:

Shortest distance(perpendicular distance) of a point (α,β) from a line ax + by + c = 0 is given by

$d \: = \frac{ |a \alpha + b \beta + c| }{ \sqrt{ {a}^{2} + {b}^{2} } }$

Using this

d = |-2*5+3*6+4|/√(2² + 3²)

= 12/√13