Question

Find k if the points are collinear (2,1),(k,-1) and (-1,3) ​

Answer 1

As the given points are collinear that means they lies on a single line it implies that area of triangle formed by joining these points must be equals to zero and by putting this restriction on these three points we can easily calculate the value of k by putting the area of triangle equals to zero as shown in attachment.

Answer 2

Solution:

Given points:

=> A(2, 1)

=> B(k, -1)

=> C(-1, 3)

To Find:

=> Value of k

Formula used:

$\sf{\implies Area\;of\;\triangle\;ABC= \dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]=0}$

So,

$\sf{\implies Area\;of\;\triangle\;ABC= \dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]=0}$

Where,

=> $x_{1}$ = 2

=> $y_{1}$ = 1

=> $x_{2}$ = k

=> $y_{2}$ = -1

=> $x_{3}$ = -1

=> $y_{3}$ = 3

Put the values in the formula,

$\sf{\implies \dfrac{1}{2}[2(-1-3)+k(3-1)+(-1)(1-[-1])]=0}$

$\sf{\implies \dfrac{1}{2}[2(-4)+k(2)-1(2)]=0}$

$\sf{\implies \dfrac{1}{2}[-8+2k-2]=0}$

$\sf{\implies -8+2k-2=0}$

$\sf{\implies -8+2k=2}$

$\sf{\implies 2k=10}$

$\sf{\implies k=\dfrac{10}{2}}$

$\large{\boxed{\boxed{\red{\sf{\implies k=5}}}}}$