Three vertices of a triangle are A(4, 3), B(1, −1) and C(7,k). the value of k such that the centroid orthocenter, circumcentre and incetre lie on the same line is
Answers
Answer:
Given A(4,3) B(1,-1) C(7,k) .
centroid ,orthoentre ,incentre and circumcentre of the triangle ABC lie on the same straight line then it is isosceles triangle.
so we can write AB = AC
=> square of (AB) = square of (AC)
9 + 16 = 9 + square of (k-3)
After solving
k = -1 and 7..Ans
We can also solve for taking two another side also.
Step-by-step explanation:
i thinknit is help full
ANSWER:
Given:
- Vertices of triangle = A(4,3), B(1,-1) and C(7,k)
To Find:
- Value of k such that centroid, orthocentre, circumcentre and incentre lie on same line.
Solution:
We know that, if centroid, orthocentre, circumcentre and incentre lie on same line, then the corresponding triangle is an isosceles triangle.
Hence, in ∆ABC,
⇒ AB = AC.
To calculate AB and AC, we use distance formula.
Distance formula:
Distance formula:⇒ AB = √[(x2-x1)²+(y2-y1)²]
Here, x1,y1 are coordinates of A and x2,y2 are of B.
So,
⇒ AB = √[(1-4)²+(-1-3)²]
⇒ AB = √[3²+(-4)²]
⇒ AB = √[9+16] = √[25] ------(1)
And,
⇒ AC = √[(7-4)²+(k-3)²]
⇒ AC = √[(3)²+(k-3)²]
⇒ AC = √[9+(k-3)²]------(2)
As,
⇒ AB = AC
from (1) & (2),
⇒ √[25] = √[9+(k-3)²]
Squaring both sides,
⇒ 25 = 9 + (k-3)²
We know that, (a-b)² = a²+b²-2ab. So,
⇒ 25 = 9 + k² + 9 - 6k
Transposing, 25 to RHS,
⇒ 0 = k² - 6k + 18 - 25
⇒ k² - 6k - 7 = 0
Splitting the middle term,
⇒ k² - 7k + k - 7 = 0
⇒ k(k - 7) + 1(k - 7) = 0
⇒ (k - 7)(k + 1) = 0
⇒ k = 7 or -1
So, for k = 7 & -1, the centroid, orthocentre, circumcentre and incentre of the ∆ABC, lie on same line.
Formulae Used:
- Distance formula: AB = √[(x2-x1)²+(y2-y1)²]
- (a-b)² = a²+b²-2ab.
Learn More:
- Centroid: Point of intersection of the three medians.
- Orthocentre: Point of intersection of the three heights of a triangle.
- Circumcentre: Point of intersection of the three perpendicular bisectors.
- Incentre: Point of intersection of the three angle bisectors.