Question 1111745
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?
{{{drawing(300,300,-6,6,-1,11,
blue(line(-5,1,0,4)),
green(line(-5,1,0,1)),
green(rectangle(0,1,-0.5,1.5)),
red(line(-5,1,0,8)),
blue(rectangle(0,0,5,10)),
locate(2,5.5,blue(pool)),
locate(-5.1,0.9,P),
locate(-2.5,1,d),
locate(-2.5,2.5,e),
locate(-2.5,4.5,f)
)}}} 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(x[P],y[P])}}}
and a line with equation {{{ax+by+c=0}}} as
{{{distance}}}{{{"="}}}{{{abs(ax[P]+by[P]+c)/sqrt(a^2+b^2)}}} .
 
In this case, your point is the origin {{{O(0,0)}}} ,
so {{{x[P]=y[P]=0}}} , and
{{{y=(5/13)x + 5}}} can be transformed,
by rearranging and multiplying both sides of the equal sign time {{{13}}} ,
to get the equivalent equation
{{{5x-13y+65=0}}} , with {{{a=5}}}, {{{b=-13}}} , {{{c=65}}} .
 
So, in this case,
{{{distance}}}{{{"="}}}{{{65/sqrt(5^2+(-13)^2)}}}}{{{"="}}}{{{65/sqrt(25+169)}}}}{{{"="}}}{{{65/sqrt(194)}}} .
The elegant way to express the exact value is
{{{distance=65sqrt(194)/194}}}
The approximate value for that irrational number
is the never ending, non-repeating, decima {{{"l4.666728...."}}} .
 
ANOTHER APPROACH:
It is easy to see that the x- and y-intercepts are the points
{{{A(-13,0)}}} and {{{B(0,5)}}} .
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:
{{{drawing(600,300,-16,4,-3,8,grid(0),
green(triangle(0,0,-1.675,4.356,0,5)),
line(-16,-1.154,5,6.923),locate(0.1,0,O)
locate(-13.2,0.6,A),locate(0.1,5,B),
locate(-1.85,4.9,C),locate(-1.3,3.9,green(d))
)}}}
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(13^2+5^2)=sqrt(169+25)=sqrt(194)}}} ,
calculated using the Pythagorean theorem.
The area of ABO is {{{area=5*13/2}}} ,
calculated as {{{area=base*height/2}}} ,
using AO as the base and OB as the height.
As we know that the distance {{{green(d)}}}
is measured along line OC, perpendicular to AB,
we could also calculate the area as {{{area=AB*d/2}}} ,
using {{{AB}}} as the base and {{{green(d)=OC}}} as the height.
Substituting, and combining both ways of calculating area
{{{sqrt(194)d/2=5*13/2}}}
{{{sqrt(194)d=5*13}}}
{{{d=5*13/sqrt(194)=65/sqrt(194)=65sqrt(194)/194=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/13=5/sqrt(194)}}} , so
{{{d=5*13/sqrt(194)=65/sqrt(194)=65sqrt(194)/194=aproximately14.67}}} .