Question

find k is points (k+1,2k),3k,2k+3) 5k- 1,5k) are collinear​

Answer 1

Answer:

The value of k is 2

Step-by-step explanation:

Let the given points be

A(k+1, 2k), B(3k, 2k+3) and C(5k- 1,5k)

since the points A,B and are collinear,

slope of AB = slope of BC

$\boxed{\text{\bf\,Slope of the line joining }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is }\bf\frac{y_2-y_1}{x_2-x_1}}$

$\frac{2k+3-2k}{3k-(k+1)}=\frac{5k-(2k+3)}{(5k-1)-3k}$

$\implies\frac{3}{2k-1}=\frac{3k-3}{2k-1}$

$\implies\frac{1}{2k-1}=\frac{k-1}{2k-1}$

$\implies\,1=k-1$

$\implies\,\boxed{\bf\,k=2}$

Answer 2

Given:

(k+1,2k), (3k,2k+3), and (5k- 1,5k) are collinear.

To find:

k=?

Solution:

In the is question we assume the given point making triangle and its formed is =0 Then:

Formula:

$\bold{Area= \frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}$

Area= 0

$x_1= k+1\\y_1=2k\\x_2=3k\\y_2=2k+3\\x_3=5k-1\\y_3=5k$

$0=\frac{1}{2}[(k+1)((2k+3)-(5k))+3k(5k-2k)+(5k-1)(2k-(2k+3))]\\\\0= \frac{1}{2}[(k+1)(2k+3-5k)+3k(3k)+(5k-1)(2k-2k-3)]\\\\0= \frac{1}{2}[(k+1)(3-3k)+9k^2+(5k-1)(-3)]\\\\0= \frac{1}{2}[3k-3k^2+3-3k+9k^2-15k+3)]\\\\0= [6k^2-15k+6]\\\\0=3 [2k^2-5k+2]\\\\0= [2k^2-5k+2] \\\\2k^2-5k+2=0$

$2k^2-4k-1k+2=0\\\\2k(k-2)-1(k-2)=0\\\\k-2=0 \ \ \ and \ \ \ 2k-1=0\\\\k=2 \ \ \ \ and \ \ \ k= \frac{1}{2}$

The final value of k is 2.