Question

Show by section formula that the point (3,-2) (5,2) and (8,8) are collinear

Answer 1

Answer:

0. It is a collinear.

Step-by-step explanation:

(3, -2) (5, 2) (8, 8)

(x1-y1) (x2, y2) (x3, y3)

1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

=1/2[3(2-8)+5(8-(-2))+8(-2-2)

=1/2[3(-6)+5(8+2)+8(-4)]

=1/2[-18+5(10)-32]

=1/2[-18+50-32]

=1/2[32-32]

=1/2[0]

=0/2

=0

Answer 2

ANSWER:

Given:

• Three points: A(3, -2); B(5, 2); C(8, 8)

To Prove:

• These points are collinear

Concept Used:

We will take these points as vertices of a triangle and then find the area of the triangle. The area will come to be 0, which proves that the points lie on same line, i.e., are collinear.

Solution:

Let the 3 points A(3, -2), B(5, 2), C(8, 8) be the vertices of a triangle.

Now, we'll find the area of the ∆ ABC.

We know that,

$:\implies\sf Area\:of\:triangle=\dfrac{1}{2}\bigg|(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|$

Here,

$:\longrightarrow\sf x_1=3,\:x_2=5,\:x_3=8,\:y_1=-2,\:y_2=2,\:y_3=8$

So,

$:\implies\sf Area_{ABC}=\dfrac{1}{2}\bigg|(3(2-8)+5(8+2)+8(-2-2)\bigg|$

So,

$:\implies\sf Area_{ABC}=\dfrac{1}{2}\bigg|(3(-6)+5(10)+8(-4)\bigg|$

On Simplifying,

$:\implies\sf Area_{ABC}=\dfrac{1}{2}\bigg|-18+50-32\bigg|$

So,

$:\implies\sf Area_{ABC}=\dfrac{1}{2}\bigg|50-50\bigg|$

$:\implies\sf Area_{ABC}=\dfrac{1}{2}\bigg|0\bigg|$

Hence,

$:\implies\bf Area_{ABC}=0$

As, the area of ∆ABC = 0, the points lie on same line.

Therefore, the points A(3, -2), B(5, 2), C(8, 8) are collinear.

HENCE PROVED!!