Question

Find the value of $k$ if $A(4,11),B(2,5),C(6,k)$ are collinear points.

Answer 1

Given: Points $A(4,11),B(2,5)$ and $C(6,k)$ are collinear.
Value of $k$: $A(4,11)=(x_1,y_1),B(2,5)=(x_2,y_2),C(6,k)=(x_3,y_3)$
Slope of line $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{5-11}{2-\left(4\right)}=\frac{-6}{-2}=3$
$\therefore$ Slope of line $AB=3$ ........ (1)
Slope of line $BC=\frac{y_3-y_2}{x_3-x_2}=\frac{k-5}{6-2}=\frac{k-5}{4}$
$\therefore$ Slope of line $BC=\frac{k-5}{4}$ ......... (2)
$\therefore$ Slope of line $AB=$ Slope of line $BC$
$\therefore 3=\frac{k-5}{4}$ (From (1) and (2))
$\therefore \frac{k-5}{4}=3$
$\therefore k-5=12$
$k=12+5$
$=17$