Question

Find the value of K or which the poitn (2,5) and (K,11/2) and (4,6) are collinear

Answer 1

Answer:

value of k is: 3

Step-by-step explanation:

Three points A($x_{1}$,$y_{1}$), B($x_{2}$,$y_{2}$) and C($x_{3}$,$y_{3}$) are said to be collinear if:

slope of AB= slope of BC= slope of AC

Here A=(2,5), B=(K,11/2), C(4,6)

slope of AB=$\dfrac{(\dfrac{11}{2} )-5}{k-2}=\dfrac{\dfrac{1}{2}}{k-2}$

slope of AC= $\dfrac{6-5}{4-2}=\dfrac{1}{2}$

now for A,B,C to be collinear

slope of AB= slope of AC

That means   $\dfrac{\dfrac{1}{2}}{k-2}=\dfrac{1}{2}$

$k=3$

Hence, the value of k for A,B,C to be collinear is 3.

Answer 2

Answer:

Answer Is 3

Step-by-step explanation:

3 points A (x1,y1), B (x2,y2) and C (x3,y3) are the collinear If slope of AB=BC=CA

A=(2,5),B=(k,11/2),C (4,6)

AB=(11/2)-5/k-2=1/2/k-2

AC=6-5/4-2=1/2

A,B, C to be collinear AB=AC=1/2/K-2=1/2

K=3