SOLUTION: Find the perpendicular distance between the line y=(5/13)x + 5 and the origin.

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Question 1111745: Find the perpendicular distance between the line y=(5/13)x + 5 and the origin.
Found 2 solutions by math_helper, KMST:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
First draw this:


Truncated y = (5/13)x + 5 to form a triangle with the -x axis and +y axis.
d is the distance sought, and is represented by the green line.

We know the length of the leg of the triangle that goes along the y-axis is 5 (from y=(5/13)x+5, by inspection the y intercept is 5).

d = 5sin(a)
What is a? To find that we need to find (x0,0) which will tell us the angles of the large triangle.
0 = (5/13)x0 + 5 —> x0 = -13 so the length of the triangle along the -x axis is 13.

The angle a is therefore +tan%5E%28-1%29%2813%2F5%29+=+68.96248897++ degrees.
d = 5 sin(68.96248897) = +highlight%284.666728%29+ units.
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EDIT 3/10/18 — Thank you tutor KMST for your approach. I think you may have inadvertently wrote 14.6667 as the answer and carried it throughout(?). I noticed only because clearly d < 5.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Distance from a point to a line is always measured along a perpendicular, of course.
If you were at point P in the sketch below,along what path would you measure your distance to the pool?
along what path would you measure your distance to the pool, d, e, or f?

MY GUESS AS TO THE EXPECTED SOLUTION:
Your teacher (or textbook) has probably given you the formula
to calculate the distance between a point P%28x%5BP%5D%2Cy%5BP%5D%29
and a line with equation ax%2Bby%2Bc=0 as
distance%22=%22abs%28ax%5BP%5D%2Bby%5BP%5D%2Bc%29%2Fsqrt%28a%5E2%2Bb%5E2%29 .

In this case, your point is the origin O%280%2C0%29 ,
so x%5BP%5D=y%5BP%5D=0 , and
y=%285%2F13%29x+%2B+5 can be transformed,
by rearranging and multiplying both sides of the equal sign time 13 ,
to get the equivalent equation
5x-13y%2B65=0 , with a=5, b=-13 , c=65 .

So, in this case,
distance%22=%2265%2Fsqrt%285%5E2%2B%28-13%29%5E2%29}%22=%2265%2Fsqrt%2825%2B169%29}%22=%2265%2Fsqrt%28194%29 .
The elegant way to express the exact value is
distance=65sqrt%28194%29%2F194
The approximate value for that irrational number
is the never ending, non-repeating, decima %22l4.666728....%22 .

ANOTHER APPROACH:
It is easy to see that the x- and y-intercepts are the points
A%28-13%2C0%29 and B%280%2C5%29 .
If you look at the line, the axes,
and the path along which you would measure the distance to the origin,
you see right triangles:

When people see right triangles they think about
trigonometric ratios,
or the Pythagorean theorem,
or similar triangles.
Each of those ideas can also lead to a solution,
as another tutor showed you for trigonometric ratios.

USING THE PYTHAGOREAN THEOREM:
The most obvious right triangle in the sketch above is big triangle ABO ,
with hypotenuse AB=sqrt%2813%5E2%2B5%5E2%29=sqrt%28169%2B25%29=sqrt%28194%29 ,
calculated using the Pythagorean theorem.
The area of ABO is area=5%2A13%2F2 ,
calculated as area=base%2Aheight%2F2 ,
using AO as the base and OB as the height.
As we know that the distance green%28d%29
is measured along line OC, perpendicular to AB,
we could also calculate the area as area=AB%2Ad%2F2 ,
using AB as the base and green%28d%29=OC as the height.
Substituting, and combining both ways of calculating area
sqrt%28194%29d%2F2=5%2A13%2F2
sqrt%28194%29d=5%2A13
d=5%2A13%2Fsqrt%28194%29=65%2Fsqrt%28194%29=65sqrt%28194%29%2F194=aproximately14.67

USING SIMILAR TRIANGLES:
Triangles ABO and BCO are both right triangles,
and they both have the same angle at B,
so they are similar triangles,
with BCO being a scaled-down version of ABO.
The ratio of corresponding sides is the scale factor,
the same for the long leg as for the hypotenuse
d%2F13=5%2Fsqrt%28194%29 , so
d=5%2A13%2Fsqrt%28194%29=65%2Fsqrt%28194%29=65sqrt%28194%29%2F194=aproximately14.67 .