Question

$\huge \: \large{{\underline{\color{blue}{\bf{ Verified \: \: answer}}}}}$


(a, b+c), (b, c+a) and (c, a+b)

determine whether is sense of points are collinear​

Answer 1

Answer:

Question :

(a, b + c), (b, c + a) and (c, a + b)

determine whether is sense of points are collinear​

Solution :

We know that, if points are collinear, their slopes between any two points are equal.

Let, A(a, b + c), B(b, c + a) and C(c, a + b)

We know that,

$\boxed{\begin{array}{cc}\mathrm{If\ (x_1,y_1)\ and\ (x_2,y_2)\ are\ two\ points}\\\text{then}\\\mathrm{Slope = \dfrac{y_2-y_1}{x_2-x_1}}\end{array}}$

So, for being collinear, Slope of AB = Slope of BC

Applying the formula, we get,

$\implies \mathrm{\dfrac{(c+a) - (b+c)}{b-a} = \dfrac{(a+b)-(c+a)}{(c-b)}}$

$\implies \mathrm{\dfrac{a-b}{b-a} = \dfrac{b-c}{c-b}}$

=> -1 = -1

Here, both the sides matched

Hence proved.

Thus, these points are collinear

Hope it helps!!

Answer 2

$\large\underline{\sf{Solution-}}$

Given coordinates are (a, b+c), (b, c+a) and (c, a+b).

So, in order to check whether the points are collinear or not, we have to find the area of triangle formed by these coordinates, if area of triangle is 0, the points are collinear.

So, Let evaluate the area of triangle.

We know,

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of a triangle. Then, area of triangle ABC is given by

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

Here,

• • x₁ = a

• • x₂ = b

• • x₃ = c

• • y₁ = b + c

• • y₂ = c + a

• • y₃ = a + b

$\rm = \dfrac{1}{2}\bigg |a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)\bigg|$

$\rm = \dfrac{1}{2}\bigg |a(c - b) + b(a - c) + c(b- a)\bigg|$

$\rm = \dfrac{1}{2}\bigg |ac - ab + ba - bc+ cb- ac\bigg|$

$\rm \:  =  \: \dfrac{1}{2} \times 0$

$\rm \:  =  \: 0$

So, area of triangle = 0

$\rm \implies\:(a, b+c), (b, c+a) \: and \: (c, a+b) \: are \: collinear.$

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Learn More:

1. Section formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$