Find the value of for which the points (–5, 1), (1,) and (4, –2) are collinear.
Answers
Step-by-step explanation:
Given -
- (-5,1), (1,k), (4,-2) are collinear
To Find -
- Value of k
Method 1 :-
As we know that :-
- x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
→ (-5,1) = (x1,y1)
(1,k) = (x2,y2)
(4,-2) = (x3,y3)
Now,
→ -5(k + 2) + 1(-2 - 1) + 4(1 - k) = 0
→ -5k - 10 - 3 + 4 - 4k = 0
→ 9k = -9
→ k = -1
Method 2 :-
As we know that :-
- Slope = y2 - y1/x2 - x1
Now,
Let a = (-5,1), b = (1,k), c = (4,-2)
Then,
Slope of line AB =
(-5,1) = (x1,y1)
(1,k) = (x2, y2)
→ k - 1/1 + 5
→ k - 1/6
And
Slope of line BC =
(1,k) = (x1,y1)
(4,-2) = (x2,y2)
→ -2 - k/4 - 1
→ -2 - k/3
Now,
→ k - 1/6 = -2 - k/3
→ 3(k - 1) = 6(-2 - k)
→ 3k - 3 = -12 - 6k
→ 9k = -12 + 3
→ 9k = -9
→ k = -1
Hence,
The value of k is -1
- (–5, 1), (1, k) and (4, –2) are collinear.
- value of k ?
- x1(y2 - y3) + x2 (y3 - y1) + x3(y1 - y2)
x1 = -5. y1 = 1
x2 = 1. y2 = k
x3 = 4. y3 = -3
-5(k - 2) + 1(-2 - 1) + 4(1 - k) = o
-5k - 10 - 3 + 4 - 4k = 0
9k = -9
k = -1
slope =y2 - y1 / x2 - x1
Let a = (-5, 1), b = (1, k), c = (4, -2)
(-5, 1) = (x1, y1)
(1, - k) = (x2 - x1)
k = 1/1 + 5
k = 1/6
(1, k) = (x2, y2)
(4, - 2) = (x3 - x3)
-2 - k/4 - 1
-2 - k/3
k - 1/6 = -2 - k/3
3(k - 1) = 6(-2 - k)
3k - 3 = -12 - 6k
9k = -12 + 3
9k = -9
k = -1