Find the distance between \((3,4)\) and the line \(4x+3y+7=0\).
Any help is greatly appreciated.
thx in advance!
Here is one way:
4x+3y=-7
3y = -4x-7
y = -4/3 x- 7/3 so the slope is -4/3 Perpindicular slope is 3/4
Find line with 3/4 slope and point 3,4
4 = 3/4 (3) + b
b= 7/4
y = 3/4 x + 7/4
Now equate the two lines:
3/4x + 7/4 = -4/3 x - 7/3
9/12 x + 7/4 = - 16/12 x - 7/3
25/12x = - 28/12 - 21/12
25/12 x = -49/12
x = -49/25
y= 196/75 - 7/3 = 196-7(25) /75 = 7/25
Now use distance formual between 3,4 and -49/12 , 7/25
d^2 = (3 - - 49/25)^2 + (4 - 7/25)^2
= 124/25 ^2 + 93/25 ^2 = 24025/625 take the square root = 155/25 = 6.2
Here is another way:
https://brilliant.org/wiki/dot-product-distance-between-point-and-a-line/
Here's a quick way......the distance between a point (a,b) and a line in the form of
Ax + By + C = 0 =
abs (A(a) + B(b) + C)
_________________ so we have
sqrt (A^2 + B^2 )
abs ( 4(3) + 3(4) + 7 ) 31 31
___________________ = ______ = ___ = 6.2 units
sqrt (4^2 + 3^2 ) sqrt(25) 5