Question

If the points P, Q(x, 7), R, S(6, y) in this order divide the line segment joining A(2, p) and B(7, 10) in 5 equal parts, find x, y and p. [CBSE 2015]

Answer 1

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 = $\frac{1}{5}$AB .....(2)

Now,

AS = AP + PQ + QR + RS = $\frac{1}{5}$AB +$\frac{1}{5}$AB + $\frac{1}{5}$AB + $\frac{1}{5}$AB = $\frac{4}{5}$AB .....(3)

From (2) and (3), we get

AS : SB = $\frac{4}{5}$AB : $\frac{1}{5}$AB = 4 : 1

Similarly,

AQ : QB = 2 : 3

Using section formula, we get

Coordinates of Q = $\left(\frac{2\times 7+3\times 2}{2+3},\frac{2\times 10+3\times p}{2+3}\right)=\left(\frac{20}{5},\frac{20+3p}{5}\right)=\left(4,\frac{20+3p}{5}\right)$

$\therefore \left(x,7\right)=\left(4,\frac{20+3p}{5}\right)$

$\Rightarrow x=4$ and $7=\frac{20+3p}{5}$

Now,

$$\begin{gathered} 7=\frac{20+3p}{5} \\ \Rightarrow 20+3p=35 \\ \Rightarrow 3p=15 \\ \Rightarrow p=5 \end{gathered}$$

Coordinates of S = $\left(\frac{4\times 7+1\times 2}{4+1},\frac{4\times 10+1\times p}{4+1}\right)=\left(\frac{30}{5},\frac{40+5}{5}\right)=\left(6,9\right)$

$$\begin{gathered} \therefore \left(6,y\right)=\left(6,9\right) \\ \Rightarrow y=9 \end{gathered}$$

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