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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

\[A\left( x_1 , y_1 \right)\] \[P\left( x_2 , y_2 \right)\], \[R\left( x_3 , y_3 \right)\] and B \[\left( x_4 , y_4 \right)\].

QR = RS

\[R\left( x_3 , y_3 \right) = \left( \frac{12 + 4}{2}, \frac{14 + 18}{2} \right) = \left( 8, 16 \right)\]

RS = SB

\[B\left( x_4 , y_4 \right) = \frac{8 + x_4}{2} = 4, \frac{16 + y_4}{2} = 18\]

\[ \Rightarrow 8 + x_4 = 8, y_4 = 36 - 16\]

\[ \Rightarrow x_4 = 0, y_4 = 20\]

\[B\left( x_4 , y_4 \right) = \left( 0, 20 \right)\]

\[\frac{AQ}{QB} = \frac{AP + PQ}{QR + RS + SB} = \frac{AP + AP}{AP + AP + AP} = \frac{2}{3}\]

\[\frac{2 \times 0 + 3 \times x_1}{2 + 3} = 12, \frac{2 \times 20 + 3 \times y_1}{2 + 3} = 14\]

\[ \Rightarrow 3 x_1 = 60, 40 + 3 y_1 = 14\]

\[ \Rightarrow x_1 = 20, y_1 = 10\]

\[A\left( x_1 , y_1 \right)\] = (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)