A(2, -1), B(-5, 3) and C(-7, k) are collinear. Find the value of ‘k’=…………..
Answers
Step-by-step explanation:
Given :-
A(2, -1), B(-5, 3) and C(-7, k) are collinear
To find:-
Find the value of k ?
Solution:-
Given points are : A(2, -1), B(-5, 3) and C(-7, k)
Let (x1, y1)=(2,-1)=>x1=2 and y1 = -1
Let (x2, y2)=(-5,3)=>x2=-5 and y2=3
Let (x3, y3)=(-7,k)=>x3 = -7 and y3 = k
Given that the points are Collinear points.
We know that
If the given points are collinear points then the area of the triangle formed by the given points is zero.
Area of a triangle formed by the points (x1, y1),
(x2, y2) and (x3, y3) is
(1/2)| x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
We have
(1/2)| x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| = 0
On Substituting these values in the above formula
=> (1/2) | 2(3-k) +(-5)(k+1)+(-7)(-1-3) | = 0
=> (1/2) | 6-2k-5k-5+(-7)(-4) | = 0
=> (1/2) | 6-2k-5k-5+28 | = 0
=> (1/2) | 29-7k | = 0
=> | 29-7k | = 0×(2/1)
=> 29-7k = 0
=>29 = 7k
=> 7k = 29
=>k = 29/7
Therefore,k = 29/7
Answer:-
The value of k for the given problem is 29/7
Used formulae:-
- The points lies on a line are called Collinear Points.
- If the given points are collinear points then the area of the triangle formed by the given points is zero.
- Area of a triangle formed by the points are (x1, y1),(x2, y2) and (x3, y3) is
- (1/2)| x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|