SOLUTION: what is the euclidean distance from the point (3,5) to the line y=2x

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Question 274663: what is the euclidean distance from the point (3,5) to the line y=2x
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Finding a distance between a point given by coordinates (x, y) and a line given by equation y=ax+b
We want to find the perpendicular distance between a point given by coordinates (3,5)
and a line given by equation y=2%2Ax%2B0

First, let's draw a diagram of general situation with point P (xo, yo) and
line L: y= a.x + b. The required distance is PC. (in the diagram below)





Methodology
We will first find the vertices of the triangle in order to get the side lengths and then by applying
Sine Rule on right angle triangle PAB and PBC we will calculate the desired distance PC.


Step1
Calculation of the vertices of triangle PAB:

Draw a vertical line passing through the point 'P'. This line x=3 will cut the given line 'L'
at point 'A'. The X coordinate of A(x1) will be same as xo=3. To find the Y-coordinate of
'A' we will use the fact that point 'A' lies on the given line 'L' and satisfies the equation
of the line 'L'
.
Now, plug this x1=3 in to the equation of line: y=2*x+0
y1=2%2A3+%2B0
y1=6

Hence, Point (A)(x1=3,y1=6)


Similarly,
Draw a horizontal line passing through the point 'P'. This line y=5 will cut the given line 'L'
at point 'B'. The Y coordinate of B(y2) will be same as yo=5. To find the X-coordinate of
B we will use the fact that point 'B' lies on the given line 'L' and satisfies the equation
of the line 'L'
.
Now, plug this y2=5 in to the equation of line: y=2*x+0
5=2%2Ax2%2B0
x2=+%285-0%29%2F2
x2=2.5

Hence, Point (B)(x2=2.5,y2=5)


Now, we have all the vertices of the triangle PAB


Step2
Calculation of the side lengths using distance formula:

d=sqrt%28%28x1-x2%29%5E2+%2B+%28y1-y2%29%5E2%29


Hence, The side lengths PA, PB and AB are
PA=1
PB=0.5
AB=1.11803398874989


Step3
Apply Sine rule on common angle B in triangle PAB and triangle PBC.
Both triangle PAB and triangle PBC are right angle triangle and points 'A', 'B' and 'C' lay on the given line L.

Sin%28B%29=+AP%2FAB=PC%2FBP

PC=%28AP%2ABP%29%2FAB=+0.447213595499958


PC is the required perpendicular distance of the point P (3, 5) from line given
lineL1: y=2*x+0.


For better understanding of this concept, look at the Lesson based on the above concept.
Lesson




So the distance is approximately 0.447 units.