Question

2) Determinant Area of the triangle
Find area of the triangle when the vertical
is given by [4, 1] [-1, -2] [ -6, -5] are
colinear or not ?​

Answer 1

Answer:

there is no going back

Step-by-step explanation:

ANSWER

Since (3,a) lie on the line 2x - 3y = 5 ...(1)

Substituting x=3 and y=a in (1) we get

2(3) - 3(a) = 5

6 - 3a = 5

⇒3a=1

⇒a=1/3...............................

Answer 2

Answer:-

We have to prove that:

Area of the triangle formed by the vertices (4 , 1) , ( - 1 , - 2) & ( - 6 , - 5) = 0

[ If three points are collinear then area of the triangle formed by them is zero]

We know that,

$\sf Area \:of\:a\: triangle= \dfrac{1}{2} \begin{vmatrix} \sf x_1 - x_2 & \sf x_1 - x_3 \\\\\sf y_1 - y_2 & \sf y_1 - y_3 \end{vmatrix}$

Let,

• x₁ = 4

• x₂ = - 1

• x₃ = - 6

• y₁ = 1

• y₂ = - 2

• y₃ = - 5

So,

$\implies \sf \: \frac{1}{2} \begin{vmatrix} \sf 4 - ( - 1)& \sf \: 4 - ( - 6) \\ \\ \sf 1 - ( - 2) \sf & \sf 1 - ( - 5) \end{vmatrix} = 0 \\ \\ \\ \implies \sf \:\begin{vmatrix} \sf 4 + 1& \sf \: 4 + 6\\ \\ \sf 1 + 2 \sf & \sf 1 + 5 \end{vmatrix} = 0 \times 2 \\ \\ \\ \implies \sf \:\begin{vmatrix} \sf 5& \sf \: 10\\ \\ \sf 3\sf & \sf 6 \end{vmatrix} = 0 \\\\ \\ \implies \sf \: |(5)(6) - (3)(10)| = 0 \\ \\ \\ \implies \sf \: |30 - 30| = 0 \\ \\ \\ \implies \sf \: 0 = 0$

Hence, proved.