If the area of triangle is 7 sq.unit then find the value ok k by using determinant method if the vertices of triangle are
A(4,9),B(K, 0) and C(4,K)
Answers
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✤ Required Answer:
K = 2,11
✒ Given that :
Area of triangle is 7 sq.unit
vertices of triangle are
A(4,9),B(K, 0) and C(4,K)
✒ we hv to find:
value of k by using determinant method
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✤ How to Solve?
Firstly u HV to know that determinant is the scalar value which is computed from different elements of a square matrix that has certain properties of a linear transformation. Let us now learn how to use the determinant to find the area of a triangle. Let’s say that (x1, y1), (x2, y2 ), and ( x3, y3 ) are three points of the triangle in the cartesian plane. Now the area of the triangle of the will be given as:
Here, k is the area of the triangle using determinant and the vertices of the triangle are represented by (x1, y1), (x2, y2 ), and ( x3, y3 ).
U should also know how to solve quadratic equation..
ツ Now let's solve the question by our knowledge.
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✤ Solution:
According to formula,
7 = ½ [4 ( 0-K ) + K ( K - 9 ) + 4 (9 - 0)]
7 = ½ [ -4K + K^2 - 9K +36 ]
14 = K^2 - 13K +36
K^2 -13K +22 = 0
So, Let's solve the quadratic equation,
Factoring k2-13k+22
The first term is, k2 its coefficient is 1 .
The middle term is, -13k its coefficient is -13 .
The last term, "the constant", is +22
Step-1 : Multiply the coefficient of the first term by the constant 1 • 22 = 22
Step-2 : Find two factors of 22 whose sum equals the coefficient of the middle term, which is -13 .
-22 + -1 = -23
-11 + -2 = -13 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -11 and -2
k2 - 11k - 2k - 22
Step-4 : Add up the first 2 terms, pulling out like factors :
k • (k-11)
Add up the last 2 terms, pulling out common factors :
2 • (k-11)
Step-5 : Add up the four terms of step 4 :
(k-2) • (k-11)
Which is the desired factorization
So K = 2,11
☀️ Hence, solved !!
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