If the points A(2,3) , B(4,k) , C(6,-3) are collinear , find the value of k.
Answers
Step-by-step explanation:
Given:-
The points A(2,3) , B(4,k) , C(6,-3)
To find:-
Find the value of k the points A(2,3) , B(4,k) , C(6,-3) are collinear?
Solution:-
Given points are A(2,3) , B(4,k) , C(6,-3)
Let (x1, y1)=(2,3)=>x1=2 and y1 = 3
Let (x2, y2)=(4,k)=>x2=4 and y2= k
Let (x3, y3)=(6,-3)=>x3= 6 and y3= -3
We know that
If the points (x1, y1),(x2, y2) and (x3, y3) are collinear points then the area of the triangle formed by the points is zero.
Area of the triangle =
∆=(1/2) | x1(y2-y3+x2(y3-y1)+x3(y1-y2) | sq.units
Given that
The points are Collinear points
=> Area of the triangle formed by them = 0
∆=(1/2) | x1(y2-y3+x2(y3-y1)+x3(y1-y2) | = 0
=> (1/2) | 2(k-(-3))+4(-3-3)+6(3-k) | = 0
=>(1/2) | 2(k+3)+4(-6)+6(3-k) |= 0
=>(1/2) | 2k+6-24+18-6k | = 0
=> (1/2) | 2k-6k-24+24 | = 0
=> (1/2) | -4 k+0| = 0
=> (1/2)| -4k|=0
=> 4k/2 = 0
=> 2k = 0
=> k = 0
The value of k = 0
Answer:-
The value of k for the given problem is 0
Used formula:-
If the points (x1, y1),(x2, y2) and (x3, y3) are collinear points then the area of the triangle formed by the points is zero.
Area of the triangle =
∆=(1/2) | x1(y2-y3+x2(y3-y1)+x3(y1-y2) | sq.units