The position vectors of point A,B and C are 9i-10j , 4i+2j and Ki-2j respectively.Find the value of k if the point A,B and c are collinear. Help me solve this.thank u

What is the vector from A to B? What is the vector from A to C? Do you know what "colinear" means?

Do you know what collinear means?

Hello, sharmala1

The position vectors of points A, B and C are 9i-10j , 4i+2j and Ki-2j respectively.
Find the value of k if the points A,B and c are collinear.

$\overrightarrow{BA} \:=\:\langle 5,\,\text{-}12\rangle$

$\overrightarrow{BC} \:=\:\langle (K\!-\!4),\,\text{-}4\rangle$

If $A,B,C$ are collinear, then: . $\overrightarrow{BA} \parallel \overrightarrow{BC}$

$\text{That is: }\:a\cdot\overrightarrow{BC} \:=\:\overrightarrow{BA}\,\text{ for some real number }a \ne0.$

We have: . $a\cdot\langle(K\!-\!4),\,\text{-}4\rangle \;=\;\langle5,\,\text{-}12\rangle$

. . . . . . . . . $\langle a(K\!-\!4),\,\text{-}4a\rangle \;=\;\langle 5,\,\text{-}12\rangle$

Hence: . $\begin{Bmatrix}a(K\!-\!4) &=& 5 & [1] \\ \text{-}4a &=& \text{-}12 & [2] \end{Bmatrix}$

From [2]: . $a\,=\,3$

Subtitute into [1]: . $3(K-4) \:=\:5 \quad\Rightarrow\quad K-4\:=\:\tfrac{5}{3}$

Therefore: . $K \:=\:\frac{17}{3}$

Equating the gradients can also be done.
(9,-10),(4,2),(k,-2)

$m=\frac{2-(-10)}{4-9}=\frac{-2-2}{k-4}$

and solve for k to get what Soroban got.