The line segment AB is divided into five congruent parts at P, Q, R and S such that A-P-Q-R-S-B. If point Q(12, 14) and S(4, 18) are given find the coordinates of A, P, R,B.
19«. The line segment AB is divided into five congruent parts at P, Q, R and S such that A-P-Q-R-S-B. If point Q(12, 14) and S(4, 18) are given find the coordinates of A, P, R,B.
Solution
Let the coordinates be
[Aleft( x_1 , y_1 right)] [Pleft( x_2 , y_2 right)], [Rleft( x_3 , y_3 right)] and B [left( x_4 , y_4 right)].
QR = RS
[Rleft( x_3 , y_3 right) = left( frac{12 + 4}{2}, frac{14 + 18}{2} right) = left( 8, 16 right)]
RS = SB
[Bleft( x_4 , y_4 right) = frac{8 + x_4}{2} = 4, frac{16 + y_4}{2} = 18]
[ Rightarrow 8 + x_4 = 8, y_4 = 36 – 16]
[ Rightarrow x_4 = 0, y_4 = 20]
[Bleft( x_4 , y_4 right) = left( 0, 20 right)]
[frac{AQ}{QB} = frac{AP + PQ}{QR + RS + SB} = frac{AP + AP}{AP + AP + AP} = frac{2}{3}]
[frac{2 times 0 + 3 times x_1}{2 + 3} = 12, frac{2 times 20 + 3 times y_1}{2 + 3} = 14]
[ Rightarrow 3 x_1 = 60, 40 + 3 y_1 = 14]
[ Rightarrow x_1 = 20, y_1 = 10]
[Aleft( x_1 , y_1 right)] = (20, 10)
AP = PQ
[frac{20 + 12}{2} = x_2 , frac{10 + 14}{2} = y_2 ]
[ Rightarrow frac{32}{2} = x_2 , frac{24}{2} = y_2 ]
[ Rightarrow x_2 = 16, y_2 = 12]
Explanation:-
Given the coordinates of four points A, P, R and B, we need to find the coordinates of point P and verify if AP = PQ.
First, we find the coordinates of point R using the midpoint formula. We are given QR = RS, so we can find the coordinates of R as follows:
R = ((Q+R)/2) = ((12+4)/2, (14+18)/2) = (8, 16)
Next, we find the coordinates of point B using the fact that RS = SB. Let B = (x4, y4). Then,
x4 = (8+x4)/2 => x4 = 0 y4 = (16+y4)/2 => y4 = 20
So, B = (0, 20).
Now, we can find the ratio of AQ to QB using the given equation:
(AQ/QB) = (AP+PQ)/(QR+RS+SB) = (AP+AP)/(AP+AP+AP) = 2/3
Substituting the coordinates of A and B, we get:
(20 + 3x1)/(2+3) = 12 (220 + 3y1)/(2+3) = 14
Solving for x1 and y1, we get:
x1 = 20, y1 = 10
So, A = (20, 10).
Finally, to find the coordinates of point P, we use the fact that AP = PQ. Let P = (x2, y2). Then,
(x1 + x2)/2 = (20+x2)/2 = (8+4)/2 = 6 => x2 = 16 (y1 + y2)/2 = (10+y2)/2 = (14+10)/2 = 12 => y2 = 12
So, P = (16, 12).
Therefore, the coordinates of point P are (16, 12) and AP = PQ.
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)