Question

Plz solve 14question........

Answer 1

Let the given points be A(k,2k), B(3k,3k), C(3,1).

Here x1 = k, x2 = 3k, x3 = 3.

y1 = 2k, y2 = 3k, y3 = 1.

Given that the points A,B,C are collinear.

Area of triangle ABC = 0

$= > \frac{1}{2}[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 0$

$= > \frac{1}{2}[k(3k - 1) + 3k(1 - 2k) + 3(2k-3k)]=0$

= > k(3k - 1) + 3k(1 - 2k) + 3(-k) = 0

= > 3k^2 - k + 3k - 6k^2 - 3k = 0

= > -3k^2 - k = 0

= > -3k^2 = k

= > k = -1/3.

Hope this helps!