Question

if the points P,Q(x,7),R,S(6,y) in this order divide the line segment joining A(2,p) andB(7,10) in 5 equal parts find x,y,and p.​

Answer 1

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It is given that P, Q(x, 7), R, S(6, y) divides the line segment joining A(2, p) and B(7, 10) in 5 equal parts.

∴ AP = PQ = QR = RS = SB .....(1)

Now,

AP + PQ + QR + RS + SB = AB

⇒ SB + SB + SB + SB + SB = AB [From (1)]

⇒ 5SB = AB

⇒ SB = 1/5 AB .....(2)

Now,

AS = AP + PQ + QR + RS = 1/5 AB + 1/5AB + 1/5AB + 1/5 AB =4/5AB .....(3)

From (2) and (3), we get

AS : SB = 4/5AB : 1/5 AB = 4 : 1

Similarly,

AQ : QB = 2 : 3

Using section formula, we get

Coordinates of Q =

(2*7 + 3*2)/2+3 , (2*10 + 3*p)/2+3 =

(x, 7) = (4, 20 + 3p / 5)

x = 4 and 7 = (20 + 3p) /5

Now,

7 = (20 + 3p) /5

35 = 20 + 3p

3p = 35 - 20

3p = 15

p 15/3

p = 5

Coordinates of S =

(4*7 + 1*2)/4 +1,(4*10 + p)/4 + 1=

(6,y) = (6,9) = y = 9

Thus, the values of x, y and p are 4, 9 and 5, respectively.