Division of Line Segment?
I was studying for my math exam can you help me answer this for my review in my exam next week.... i just find it difficult so can you help me..... pls.......... help me..........
A point P(x,y) is on the line through A(-4, 4) and B(5,2).
a)The coordinates of P given that the segment AB is extended throught B to P so that P is twice as far from B, and
b) The coordinates of P given that AB is extended through A to P so that P is three times as far from B as from A.
- 1 decade agoFavorite Answer
For a): I presume you mean that Point P is twice as far from A as from B. If this is so, then Point B becomes the midpoint of line segment ABP. Label the coordinates as (x_A, y_A) for A, (x_B,Y_B) for B, and (x_P,y_P) for P. Then by the midpoint formula:
x_B = (x_A + x_P)/2 (Eq1)
y_B = (y_A + y_P)/2. (Eq2)
Since A has coordinates (-4,4) and B has coordinates (5,2) then by Eq1
5 = (-4 + x_P)/2 (Eq3)
and by Eq2
2 = (4 + y_P)/2. (Eq4)
Solving Eq3 gives x_P = 14; Eq4 gives y_P = 0.
Hence, the coordinates of P are (14,0). So the line is determined by the points A(-4,4), B(5,2) and P(14,0).
For b): Point A becomes the point one-third (1/3) the way from Point P to Point B, i.e. |PA| = (1/3) |PB| or |PB|=3|PA|. The general formula for the point M(x_M,y_M) in between two points, say C(x_1,y_1) and D(x_2,y_2), that is a fraction p of the distance from C to D has coordinates
x_M = (1-p)x_C + px_D
y_M = (1-p)y_C + py_D.
In the given problem, Point P is C, B is D, and A is M, i.e.,
x_A = (1-p)x_P + px_B
y_A = (1-p)y_P + py_B.
Plugging-in the given p=1/3, A(-4,4), and B(5,2) in the preceding equations give
-4 = (2/3)x_p + (1/3)*5 (Eq5)
4 = (2/3)y_P + (1/3)*2. (Eq6)
Solving Eq5 and Eq6 for x_P and y_P, respectively, gives
x_P = ((-4)(3) - 5)/2 = -17/2 and
y_P = ((4)(3) - 2)/2 = 5.
Hence, the desired coordinates for P are (-17/2 , 5).
- modulo_functionLv 71 decade ago
The best way to solve problems like this is to first draw a little sketch and label it with the stated relationships. You can then usually see the best approach.
Distances along that line are lengths of the diagonal of various triangles given by the differences in x and y coords.
Go ahead, sketch it and you'll see what to do.