Case-Study 2: Read the following passage and answer the questions given below.
A farmer has a field in the shape ofquadrilateral, where he grews differentcrops in the shapeof triangle. For this,he needs to measure the area of field,which is given below.
Area of a triangle whose vertices are(x₁, y₁ ) , (x₂ , y₂ ) and (x₃ , y₃ ) is given bythe determinant

Since, area is a positive quantity, so wealways take the absolute value of thedeterminantD. Also, the area of thetriangle formed by three collinear pointsis zero.
Based on the above information, answerthe following questions.
(i) Find the area of the triangle whose vertices are (− 1, 2), (2, 3) and (6, 1).
(ii) If the points (2, − 3), (k, − 1) and (0, 4)are collinear, then find the value of k.
(iii) If the area of a triangle ABC, with vertices A(1, 3), B(0, 0) and C(k, 0) is 3 sq. units, then find value of k.

Question

Case-Study 2: Read the following passage and answer the questions given below.
A farmer has a field in the shape ofquadrilateral, where he grews differentcrops in the shapeof triangle. For this,he needs to measure the area of field,which is given below.
Area of a triangle whose vertices are(x₁, y₁ ) , (x₂ , y₂ ) and (x₃ , y₃ ) is given bythe determinant

Since, area is a positive quantity, so wealways take the absolute value of thedeterminantD. Also, the area of thetriangle formed by three collinear pointsis zero.
Based on the above information, answerthe following questions.
(i) Find the area of the triangle whose vertices are (− 1, 2), (2, 3) and (6, 1).
(ii) If the points (2, − 3), (k, − 1) and (0, 4)are collinear, then find the value of k.
(iii) If the area of a triangle ABC, with vertices A(1, 3), B(0, 0) and C(k, 0) is 3 sq. units, then find value of k.

Accepted Answer

(i) Area of triangle, is given by the determinanat ∆=12-121231611     =12-1(-1)-2(2-6)+(2-18)                                expanding along R1     =12-2+8-16     =12-10=5 sq units (ii)  ∵ Given points are collinear       ∵ 122-31k-11041=0        Expanding along R1, we get          2(-1-4)+3(k)+1(4k)=0          ⇒                        7k-1=0 ⇒ k=107 (iii)  According to the question,        12131001k01=±3    ⇒   131001k01 =±6    ⇒ 1(0-0)-3(0-k)+1(0-0)=±6    ⇒                                           3k=±6    ⇒                                             k=±2

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