Given j(-2,5) and k(2,-3). Find the point p on jk such that 4* jp=pk

Question

I would like to know how to solve it or a detailed answer so i cn use it to answer the rest of my questions.

Answer 1

I thought the midpoint formula would work but it doesn't. It would work if 3*jp = pk 
 
Let's try the distance formula.
 
Distance d between points j and k is sq root (16 + 64) = 4√5
 
Divide this distance by 5. Therefore the distance from point j to point p is (4/5)√5
 
The distance D is sq root (x + 2)^2 + (y - 5)^2 which must equal (4/5)√5
 
We can change this equation to one variable by using the equation of the line containing points j and k.
 
The equation is y = -2x + 1
 
So our distance equation becomes    (x + 2)^2 + (-2x + 1 - 5)^2 = 16/5
 
x^2 + 4x + 4 + 4x^2 + 16x + 16 = 16/5
 
5x^2 + 20x + 20 = 16/5
 
x^2 + 4x + 4 = 16/25
 
(x + 2)^2 = 16/25
 
x + 2 = 4/5           we ignore the negative root because point p would not lie between j and k
 
x = (4/5) - 2
 
x is approx. -1.2
y is approx. 3.4

Answer 2

Think of j , k and p as vectors and introduce a scalar called λ.   Then a general point on the line joining the points j and k can be written as the vector equation

 

p =  j +  λ (k - j)       (so when λ = 0 we have p = j and when λ =1 we have p = k )

 

Your condition on λ is   4 (p dot j)  =  p dot k       (p dot j means the dot product, or inner product between p and j)

 

The dot product is a scalar so the equation is a scalar equation for λ.   Substituting we get

 

4 [ (j dot j ) + λ (k dot j) - λ (j dot j) ] = (k dot j)  + λ  (k dot k) - λ (k dot j)

 

j dot j =  41        k dot k = 13   and  j dot k =  -19

 

with these substitutions and some algebra  we get   λ = 183 / 272

 

So p =  ( 1-  183/272 ) j   + 183/272 k

 

 

 

 

Answer 3

     Are you sure that you typed in this problem correctly? The answer to this involves coordinates with decimals. Our goal is to split JK into 5 congruent segments, that way if you take the first point closest to J and call it P, from P to J is one of the congruent segments while P to K is 4 of these congruent segments thereby satisfying your condition.

     I am using geometry to do this, because the formulas you are taught in coordinate geometry won't help you. If you plot your point J and draw a vertical segment going down, then plot point K and draw a horizontal to the left you will form a triangle. let's call where those 2 segments meet pt A having coordinates (-2,-3). You can now just count the boxes to find the length of JA and KA. JA = 8 and KA = 4. If you split each of thes into 5 congruent segments, the pieces would have a lenth of 8/5 = 1.6 and 4/5= .8 If you now label these starting from point A and going to pt J by adding 1.6 to the y value you'd get

(-2,-1.4), (-2, .2), (-2, 1.8), (-2, 3.4) and then finally pt J . Notice these 4 pts split JA into 5 congruent segments.

 

Doing the same for KA using .8 changing the x value this time, you'd get 

(-1.2, -3), (-.4,-3), (.4,-3), (1.2,-3) and then finally pt K. Notice these 4 pts split KA into 5 congruent segments.

 

Now if you draw horizontals from each of the pts on JA and verticals up from each of the points on KA, you'll see that they intersect on JK. These coordinates will split JK into 5 congruent segments. They are 

(-1.2, 3.4) (-.4, 1.8), (.4, .2), (1.2, -1.4). These 4 pts split JK into 5 congruent segments. Point P must be the closest one to point J.

 

P(-1.2,3.4) is your answer. Like I said at the start, I think I just answered a much harder question than was intended.

Hope this helped.