Question

Find the value of K for which the points (3k–1,k–2),(K, K–2),(K–1,–k–2) & (K–1,–k–2)are collinear

Answer 1

Answer:

k = 0

Step-by-step explanation:

Concept:

If the points A,B and C are collinear then

slope of AB = slope of BC

Let the given points be

A(3k–1,k–2), B(K, K–2), C(K–1,–k–2)

slope of AB

$=\frac{y_2-y_1}{x_2-x_1} \\\\=\frac{(k-2)-(k-2)}{k-(3k-1)} \\\\=\frac{0}{-2k+1)} \\\\=0$

slope of BC

$=\frac{y_2-y_1}{x_2-x_1} \\\\=\frac{(-k-2)-(k-2)}{(k-1)-k} \\\\=\frac{-k-2-k+2}{-1)} \\\\=\frac{-2k}{-1)} \\\\=2k$

since the points are collinear,

slope of AB = slope BC

0=2k

This implies

k=0