for what value of k the point (1, 4) (k -2) (-3 16) are collinear
Answers
Step-by-step explanation:
Given:-
The point (1, 4) (k -2) (-3 16)
To find:-
For what value of k the point (1, 4) ,(k, -2) ,(-3, 16) are collinear?
Solution:-
Given points are (1, 4) ,(k ,-2) ,(-3, 16)
Let (x1, y1)=(1,4)=>x1=1 and y1 = 4
Let (x2, y2)=(k,-2)=>x2=k and y2=-2
Let (x3, y3)=(-3,16)=>x3= -3 and y3=16
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) | 1(-2-16)+k(16-4)+(-3)(4-(-2)) | = 0
=> (1/2) | 1(-18)+k(12)+(-3)(4+2) | = 0
=> (1/2) | -18+12k -3(6) |=0
=> (1/2) | -18+12k-18 | = 0
=> (1/2) | 12k -36 | = 0
=> (1/2)(12k-36) = 0
=> 12k-36 = 0×2
=> 12k-36 = 0
=> 12k = 36
=> k = 36/12
=> k = 3
The value of k = 3
Answer:-
The value of k for the given problem is 3
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