Can anybody explain to me what to do?
x-2y=0 and (x+1)^2 + (y+3)^2 =5
My attempt:
sqrt5 = 2.24
find y:
x-2y=0
y=x/-2
substitute:
(x+1)^2 + (x/-2 + 3)^2 = 5
....to 1.25x^2-1x+5=0
... imaginary.
Eek. Smart people help?
Can anybody explain to me what to do?
x-2y=0 and (x+1)^2 + (y+3)^2 =5
My attempt:
sqrt5 = 2.24
find y:
x-2y=0
y=x/-2
substitute:
(x+1)^2 + (x/-2 + 3)^2 = 5
....to 1.25x^2-1x+5=0
... imaginary.
Eek. Smart people help?
Ah thank you!!!
I only have one more question for you:
For finding the distance from the point A(5,7) to the line joining B(-2,1) and C(4,-3).
Is it 8.94?
I did this by 1. Finding the midpoint of BC, then using the distance formula from (1,-1)(midpoint) to A(5,7) which equals 8.94
Is there an easier way to do this?
Because hypothetically speaking, if it were A(-4,0), and B(2,0), and C(-4,1), I cannot get the midpoint, then use the distance formula to get it correct. It would be wrong.
You can't assume the midpoint of those two points will be the shortest distance from that point to it. You could probably draw a situation in which you'll see this isn't the case. Also, it's not a line segment that you're dealing with but the entire line.
If you recall the perpendicular distance formula from a point to a line, this represents the shortest distance: $\displaystyle D = \frac{Ax_{0} + By_{0} + C}{\sqrt{A^{2} + B^{2}}}$
where $\displaystyle (x_{0},y_{0})$ is your point and $\displaystyle Ax + By + C = 0$ is the standard form of your line.
So, you find the equation of the line by finding the slope and plugging it into the point-slope equation of a line: $\displaystyle y - y_{0} = m(x-x_{0})$ and put it in the form $\displaystyle Ax + By + C = 0$ with the help of either points B or C.
Then plug it into your distance formula.
Now, I didn't plot the points out or anything so there may be a more elegant way of doing this but this works just as well.
Thanks!
Did you get 8.88?
And how do you graph this...?
There's another way to find the shortest distance by:
1. Get two perpendicular lines
2. Find out the intersects of both
3. Use distance formulas of intersections
Now. I find your way is more easier to find the distance. But how do you graph it? Last test I did, I had to stick a random line on the page.
Like:
I = line
_________
I
I_________
(-2,-1) is the only solution. The line touches the circle... its a tangent.How did you get (-2,5)?
o_O's method is the easiest way to do it. And it is derived using your idea:For finding the distance from the point A(5,7) to the line joining B(-2,1) and C(4,-3).
There's another way to find the shortest distance by:
1. Get two perpendicular lines
2. Find out the intersects of both
3. Use distance formulas of intersections
I guess my reply at: http://www.mathhelpforum.com/math-he...ne-circle.html
is out in the cold
Sorry but you're the one making the mistake. The line and circle do intersect, at the point (-2, -1).
As TheEmptySet has already stated, the graphs are not correct. In particular, the line cannot be correct - it's gradient is negative and the line x = 2y has a positive gradient.