If the vertices of the triangle axe (1,k) (4,-3) and(-9,7)
and its area is 15 square unit then find the value of K.
Answers
Step-by-step explanation:
Given :-
The vertices of the triangle axe (1,k) (4,-3) and(-9,7)
and its area is 15 square units
To find :-
Find the value of K?
Solution :-
Given vertices of a triangle are :
(1,k) (4,-3) and(-9,7)
Let (x1, y1) = (1,k) => x1 = 1 and y1 = k
Let (x2, y2)= (4,-3) => x2 = 4 and y2 = -3
Let (x3, y3) = (-9,7) => x3 = -9 and y3 = 7
We know that
Area of a triangle formed by the vertices A(x1, y1) , B(x2, y2) and C(x3, y3) is
∆ = (1/2) | x1(y2-y3) +x2(y3-y1) + x3(y1-y2) | sq.units
On Substituting these values in the above formula then
=>∆= (1/2) | 1(-3-7) +4(7-k) +(-9)(k-(-3)) | sq.units
=>∆= (1/2) | 1(-10) + 4(7-k) -9(k+3) |
=>∆= (1/2) | -10 + 28 - 4k - 9k - 27 |
=> ∆=(1/2) | (-10+28-27)+(-4k-9k) |
=>∆=(1/2) | (-9-13k) |
=> ∆=(1/2) (-13k-9)
=> ∆ = (-13k-9)/2 Sq.units
According to the given problem
Area of the given traingle = 15 sq.units
=> (-13k-9)/2 = 15
=> -13k-9 = 2×15
=> -13k -9 = 30
=> -13k = 30+9
=> -13k = 39
=> k = -39/13
=> k = -3
Therefore,k = -3
Answer:-
The value of k for the given problem is -3
Used formulae:-
Area of a triangle formed by the vertices A(x1, y1) , B(x2, y2) and C(x3, y3) is
∆ = (1/2) | x1(y2-y3) +x2(y3-y1) + x3(y1-y2) | sq.units