SOLUTION: Find the distance from (6,-1) to the line defined by y=2x-3. Please Help!!!!

Question 755490: Find the distance from (6,-1) to the line defined by y=2x-3. Please Help!!!!
Answer by Cromlix(4381) (Show Source): You can put this solution on YOUR website!
The line that is perpendicular to
y = 2x - 3 has a gradient of -1/2
For lines perpendicular to each other
their gradients multiply together to
give -1
(m1 x m2 = -1)
Using the formula y -b =m(x - a)
with m = -1/2 and coordinates (6, -1)
y - (-1) = -1/2(x - 6)
y + 1 = -1/2x + 3
y = -1/2x + 3 - 1
y = -1/2x + 2
OR 2y = -x + 4
To find where line 2y = -x + 4
and y = 2x - 3 meet, set out and
solve simultaneous equations:-
2y + x = +4....1
y -2x = -3.....2
Multiply (1) by 2
4y + 2x = 8
y - 2x = -3
Add
5y = 5
y = 1
Substitute y = 1 into Eq. 2
y - 2x = - 3
1 - 2x = - 3
- 2x = - 3 - 1
- 2x = -4
x = 2
So, this is the point where the lines meet.
Distance from (6, -1) and (1,2)
Square root((x2 - x1)^2 + (y2 - y1)^2)
Square root ( (1 - 6)^2 + (2 - (-1))^2)
Square root (5^2 + 3^2)
Square root(34)
= 5.8 units
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