Question

pls solve the 2nd one​

Answer 1

Answer:

here ia the solution of ur ques.

i hope it will help u ☺☺

Answer 2

Answer:

(ii)A(2,3) ; B(4,k) ; C(6,3)

If these points are collinear, then the area formed by the triangle by joining these points is equal to zero.

i.e., Ar. (∆ABC)=0

Ar.(∆ABC)=

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

A(2,3) ; B(4,k) ; C(6,3)

$\frac{1}{2} (2(k - 3) + 4(3 - 3) + 6(3 - k)) = 0$

$0 = (2k - 6 + 18 - 6k)$

$4k = 12$

==>k=3