- anonymous

Find the distance from the point (2,-1) to the line y = 2x + 3

- schrodinger

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- amistre64

y = -1/2 x +(2)/2 - 1
y = -1/2 x seems to be the line with the point on it
system of equation to find the intersection then distance it from point to point

- amistre64

2x+3 = -1/2 x
5/2 x = -3
x = -3/(5/2) = -6/5 maybe :)

- amistre64

-1/2 * -6/5 = 6/10 = 3/5 then
2(-6/5) + 3 = -12/5 + 15/5 = 3/5
so thats a match

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## More answers

- anonymous

My correct answer says : 8/5 * sqrt(5), could you help me to get there?

- amistre64

(10/5, -5/5)
- (-6/5 , 10/5)
---------------
(16/5)^2 + (-15/5)^2
sqrt(16^2+15^2)/5 should be it then

- amistre64

seeing how i thought my first explanation was suffieicnt; youll have to address where it is your following goes amiss at

- amistre64

other than typos :)

- amistre64

|dw:1327864212303:dw|

- amistre64

y = 2x + 3 needs a perp line to it; perp line is just flip and negate the slope
2 flips to -1/2
y = -1/2 x + b ; use (2,-1) to calibrate
2 = -1(-1)/2 + c
2 = 2 + c ; c = 0
y = -1/2 x is equation of line perp to and containing (2,-1)

- amistre64

if we know where the lines meet, we have a point of reference to measure distance with ...|dw:1327864376411:dw|

- amistre64

y = 2x + 3
-( y = -1/2 x)
-------------
0 = 5/2 x + 3 ; when x = -3/5

- amistre64

y = -1/2 * -3/5 = 3/10
therefore, point of intersection is: (-3/2 , 3/10) looks better

- amistre64

(-3/5 , 3/10) that is

- amistre64

|dw:1327864617337:dw|

- amistre64

use distance formula for distance

- amistre64

ugh!!, lost my x ... 5/2 * -2/5 * 3 = -6/5
0 = 5*-6/2*5 + 3
0 = -30/10 + 3
= -3+3
x = -6/5 ....

- amistre64

y = -1/2*-6/5 = 6/10 = 3/5
(-6/5, 3/5) is the point of intersection .....

- amistre64

\[d=\sqrt{(2-\frac{-6}{5})^2+(-1-\frac{3}{5})^2}\]
\[d=\sqrt{(\frac{10+6}{5})^2+(\frac{-5-3}{5})^2}\]
\[d=\sqrt{(\frac{16}{5})^2+(\frac{-8}{5})^2}\]
\[d=\frac{\sqrt{16^2+8^2}}{5}\]

- amistre64

that simplifies to 8sqrt(5)/5

- anonymous

|dw:1327864993315:dw|

- anonymous

that is what we found?

- amistre64

yes

- anonymous

Why doesn't using the pythagorean theorem work here?

- amistre64

it does, but first you need to know 2 points. the given point (2,-1) and the intersection of the perped lines

- amistre64

once you know the 2 points, the distance formula IS the pythag thrm

- anonymous

" the intersection of the perped lines" confused me, I am really stumped because, I still don't understand why we can't just draw the right triangle from the y axis to the point 2,-1

- amistre64

because you dont want the distance from the y axis, you want the distance to the line and the point; that distance is defined as the shortest distance between the line and the point.
which just so happens to make a 90degree angle when it hits|dw:1327865470057:dw|

- amistre64

teh y axis has nothing to do with it at all

- anonymous

OH, THE POINT OF PERPENDICULAR(ness)? to the point 2,-1?

- amistre64

yes

- anonymous

which is the part where we found the perp. point? that is the next step correct?

- anonymous

we do the negative inversse, OH, then we go from the point 2,-1 towards the other line, using the perpendicular slope, is that right?

- anonymous

until there is an intersection

- amistre64

yep

- anonymous

is there an equation to find that point?

- amistre64

its solving a system of equations;
y = 2x + 3
y = -1/2x
we can sub, eliminate, or any other method to solve the sytem

- amistre64

it turns out that i used elimination and came up with those fractions

- amistre64

y = -1/2 x
-2y = x, sub in top
y = 2(-2y) + 3
y = -4y + 3
5y = 3
y = 3/5

- amistre64

knowing y, solve for x

- amistre64

3/5 = -1/2 x
-2*3/5 = x
-6/5 = x

- amistre64

those x and y parts are the point of intersection

- anonymous

ahhh, i understand now... now i use the distance formula between the points (-6/5, 3/5) and (2,-1) Hey, i'd like to thank you for spending about an hour of your time to help me understand this, i appreciate it, much kudos to you.

- amistre64

youre welcome :)

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