Coordinate GeometryGiven three points `A(-1,2)`, `B(2,1)`, and `C(h,k)` such that the gradient of BC is twice the gradient of AC,show that `(h-k)(k-3)=6`. If C lies on the line `x-y=1` find the...
Given three points `A(-1,2)`, `B(2,1)`, and `C(h,k)` such that the gradient of BC is twice the gradient of AC,show that `(h-k)(k-3)=6`. If C lies on the line `x-y=1` find the coordinates of `C`.
First, let's start by defining that term "gradient." Usually, the word "gradient" is applied to multidimensional vector fields. In this case, though, "gradient" is simply another word for "slope." This is good because slope has a simple equation to work with:
`m = (y_2-y_1)/(x_2-x_1)`
Knowing this, we can determine the slope of lines `AC` and `BC`:
`m_(AC) = (k-2)/(h+1)`
`m_(BC) = (k-1)/(h-2)`
Now, remember how these two slopes relate. The problem says that the slope of `BC` is twice that of `AC`. This fact gives us the following relation between the above slopes:
`2m_(AC) = m_(BC)`
`2(k-2)/(h+1) = (k-1)/(h-2)`
Now, let's multiply both sides by `h+1` and `h-2` to get rid of the denominators:
`2(k-2)(h-2) = (k-1)(h+1)`
Simplifying by the FOIL method:
`2kh - 4h - 4k + 8 = kh - h + k -1`
Well, this is a mess, let's subtract the left side by the right side:
`(2kh - kh) + (-4h + h) + (-4k - k) + (8+1) = 0`
`kh - 3h - 5k + 9=0`
Huh, this does not seem to simplify to the expression we're looking for. So, let's find the function k(h) from our derived equation, and we'll compare it to the function k(h) found by solving `(h-k)(k-3) = 6` for k in terms of h.
The first equation will be:
`k_1(h) = (-9+3h)/(h-5)`
The second equation will be (a bit more complicated):
`k_2(h) = (h+3)/2 +- sqrt(-6-3h + ((3+h)/2)^2)`
Well, these don't look too similar. However, sometimes we don't see the answer just because we have not simplified in the right way algebraically. Let's compare the graphs of these two functions here.
The blue graph with the discontinuity is `k_2` and the red graph is `k_1`.
Clearly, these two are not equal, and it looks like there is a problem with the problem above!
The great part, here, is that our derived solution (`k_1`) intersects with the line `x-y = 1` while `k_2` does not. This would indicate that our solution is likely valid for the problem.
So, if C lies on the line `x-y=1` , that means that C is the point at which `x-y =1` and `k_1(h)` (graphed as `y_1(x)`) intersect.
So, we need to solve the system of equations:
Substituting for `y` and simplifying by multiplying both sides by the denominator:
`x-(-9+3x)/(x-5) = 1`
`x^2-9x+14 = 0`
We can easily factor this easily:
Giving us two intersections, where `x = 2`, and `x = 7`. Using our line equation `x-y=1` we can get the following points by solving for `y` when plugging in these values of `x`:
So this gives us 2 possible values for `C`. However, `(2,1)` is just `B`! Finding this slope would be tantamount to dividing 0 by 0, so it is not a valid possibility for `C`.
Let's check `(7,6)` to see whether `m_(BC) = 2m_(AC)`
`m_(AC) = (6-2)/(7--1) = 4/8 = 1/2`
`m_(BC) = (6-1)/(7-2) = 1`
This confirms the condition that the gradient of `BC` is twice the gradient of AC. Therefore, our answer of `(7,6)` for `C` is correct. We are also correct in saying that `(h-k)(k-3) = 6` does not characterize the point `C` . Instead the following relation characterizes the solution set for `C`:
`k = (-9+3h)/(h-5)`