Question

2. If (-8, 4). (-2,4) and (5, a) are collinear
points, then find the value of a.​

Answer 1

The value of a in the given triangle is 4.

Given :

the three vertices of the triangle is collinear (0)

The vertices are (-8,4),(-2,4) and (5,a)

To Find :

the value of a

Formula Applied :

$\text{Area of Triangle}=\frac{1}{2} ((x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3))$

Solution :

Let the vertices of the triangle as :-

A(-8,4), B(-2,4) and C(5,a)

A,B and C are vertices of the triangle in which coordinates are : $x_1=-8,y_1=4\\x_2=-2,y_2=4\\x_3=5,y_3=a$

Since we know the area of triangle = 0$(\because 0 = colinear )$

$\text{Area of Triangle}=\frac{1}{2} ((x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3))$

Hence substitute the values in formula,

$0=(((-8)(4) +(-2)(a)+(5)(4))-((4)(-2)+(4)(5)+(a)(-8)))$

$0=((-32-2a+20)-(-8+20-8a))$

$0=((-12-2a)-(12-8a))$

$0=(-12-2a-12+8a)$

$0=(-24+6a)$

$6a=24$

$a=\frac{24}{6}$

$\textbf{a = 4}$

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$\large\textbf{Explore More} :$

$\text{Area of Triangle}=\frac{1}{2} ((x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3))$

The following pictorial representation helps us to write the above formula

$\text{Area of triangle}=\frac{1}{2} \left(\begin{array}{cccc}x_1&x_2&x_3&x_1\\y_1&y_2&y_3&y_1\end{array}\right)$

$\textbf{Collinearity of three points} :$

$\implies$If three distinct points $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are colinear, then we cannot form a triangle because for such a triangle there will be no altitude(height).

$\implies$Therefore, three points $A(x_1,y_1),B(x_2,y_2)$  and $C(x_3,y_3)$will be collinear if the area Δ$ABC$ = 0.

$\implies$ Similarly, if the area of Δ$ABC$ is zero, then the three points lie on the same straight line.

$\implies$ Thus, three distinct points $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ will be collinear if and only if area of Δ$ABC=0$