find the value of k so that the area of triangle ABC with points A R K + 1, 1 b of 4 - 3 C of 7 - K is 6 square units
Answers
Question:- Find the value of k so that the area of triangle ABC with points A(K+1 , 1) , B( 4 , -3) and C(7,-k) is 6 square units
Answer: k = 3
Step-by-step explanation:
Here,
x₁ = k+1 ; y₁ = 1
x₂ = 4 ; y₂ = -3
x₃ = 7 ; y₃ = -k
We know that,
Area of Δ = 1/2([(x₁y₂ + x₂y₃ + x₃y₁) - (y₁x₂ + y₂x₃ + y₃x₁)]
6 = 1/2[(-3(k+1) - 4k + 7) - (4 - 21 - k(k+1))]
6 × 2 = [(-3k - 3 - 4k +7) - (4 - 21 - k² -k)]
12 = [4 - 7k -(-17 - k² -k )]
12 = [4 - 7k + 17 + k² + k]
12 = 21 - 6k + k²
k² - 6k + 21 - 12 = 0
k² - 6k + 9 = 0
k² - 3k - 3k + 9 = 0
k(k - 3) -3( k - 3) = 0
(k - 3)(k - 3) = 0
(k - 3)² = 0
k - 3 = 0
k = 3
Solution:-
Area of Triangle = 6 Sq. units.
Coordinates of the Points are ;
A( k+1,1) , B(4, -3) & C(7,-k).
x1 = k+1 , y1 = 1.
x2 = 4. , y2 = -3.
x3 = 7. , y3 = -k.
By Using the Area Formula, we get
Area Of ∆ ABC= 1/2 [ x1 (y2 - y3) + x2 ( y3 - y1) + x3 ( y1 - y2)]
=> 6 = 1/2 [ (k+1)(-3+k) + 4(-k -1) + 7 (1 +3)]
=> 12 = (k+1)(-3+k) + 4( -k -1) + 7(4)
=> 12 = -3k + k² -3 + k - 4k - 4 + 28
=> 12 = k² -6k + 21
=> k² - 6k +9 = 0
=> k² - ( 3 + 3)k + 9 = 0
=> k² - 3k - 3k + 9 = 0
=> K ( k - 3) -3(k-3)=0
=> ( k-3) (k-3) =0
=> K = 3.
