Question

5)
Three points (p + 1, 1), (2p + 1, 3) and (2p + 2, 2p) are collinear, if p =​

Answer 1

Answer:

$p = 2 \: \: \: \: \: \: or \: \: \: p = \frac{ - 1}{2}$

Step-by-step explanation:

Three points are collinear if their slopes are equal.

Let P=(p+1, 1)

Q=(2p+1, 3)

R=(2p+2, 2p)

Slope of PQ =

$\frac{3 - 1}{2p + 1 - p - 1} = \frac{2}{p}$

slope of QR=

$\frac{2p - 3}{2p + 2 - 2p - 1} = \frac{2p - 3}{1}$

But the slope of PQ and QR are equal.

$\frac{2}{p} = \frac{2p - 3}{1}$

${2p}^{2} - 3p = 2$

${2p}^{2} - 3p - 2 = 0$

Solving this quadratic equation by factorisation,

${2p}^{2} - 4p + p - 2 = 0$

$2p(p - 2) + 1(p - 2) = 0$

$(p - 2)(2p + 1) = 0$

$p - 2 = 0 \: \: \: \: or \: \: \: \: 2p + 1 = 0$

$p = 2 \: \: \: \: or \: \: \: p = \frac{ - 1}{2}$