if the point (2,3) (5,k) and (6,7)are collinear then prove the value of k
Answers
Step-by-step explanation:
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Step-by-step explanation:
Given :-
The point (2,3) (5,k) and (6,7) are collinear
To find :-
Find the value of k ?
Solution :-
Given points are (2,3) (5,k) and (6,7)
Let (x1, y1) = (2,3)=>x1 = 2 and y1 = 3
Let (x2, y2) = (5,k) => x2 = 5 and y2 = k
Let (x3, y3) = (6,7) => x3 = 6 and y3 = 7
Given that the points are Collinear.
We know that
If (x1, y1),(x2, y2) and (x3, y3) are collinear then the area of triangle formed by them is zero.
Area of triangle = ∆ =
(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) sq.units
Now Area of the Traingle formed by the given points is zero
=> ∆ = 0
=> (1/2)| 2(k-7)+5(7-3)+6(3-k) | = 0
=> (1/2) | 2(k-7)+5(4)+6(3-k) | = 0
=> (1/2) | 2k-14+20+18-6k | = 0
=> (1/2) | (2k-6k)+(20+18-14) | = 0
=> (1/2) | (-4k)+(38-14) | = 0
=> (1/2) | -4k+24 | = 0
=> (1/2)(-4k+24) = 0
=> (-4k+24) = 0×2
=> -4k+24 = 0
=> -4k = -24
=> 4k = 24
=> k = 24/4
=> k = 6
Therefore, k = 6
Answer:-
The value of k for the given problem is 6
Used Concept :-
→ The given points are collinear then the area of triangle formed by them is zero.
Used formulae:-
→ 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) sq.units