Question

Seg AD is a median of ∆ABC and point G is its centroid. If A ≡ (-3 , -1) and G ≡ (1 , -1). Find the coordinates of point D.

Answer 1

GiveN:-

Segment AD is a median of ∆ABC and point G is its centroid. If A(-3 , -1) and G(1 , -1).

To FinD:-

The coordinates of point D.

SolutioN:-

Analysis :

First we have to understand the question. As we are given the median of a triangle and the centroid of the median and asked to find the coordinates of one of the point of the median we will be using the mid point formula.

Solution :

Let the coordinates of point D be (x, y).

• (x₁, y₁) = A(-3, -1)
• (x₂, y₂) = D(x, y)
• (x, y) = G(1, -1)

By using mid point formula,

$\normalsize{\pink{\underline{\boxed{\bf{(x,y)=\bigg\lgroup\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup}}}}}$

The "x" of point D :

$\\ :\normalsize\implies{\sf{x=\dfrac{x_1+x_2}{2}}}$

where,

• x₁ = -3
• x₂ = x
• x = 1

Putting the values,

$\\ :\normalsize\implies{\sf{1=\dfrac{-3+y}{2}}}$

$\\ :\normalsize\implies{\sf{1\times2=-3+y}}$

$\\ :\normalsize\implies{\sf{2=-3+y}}$

$\\ :\normalsize\implies{\sf{2+3=y}}$

$\\ :\normalsize\implies{\sf{5=x}}$

$\\ \normalsize\therefore\boxed{\bf{\pink{x_2=5.}}}$

The "y" of point D :

$\\ :\normalsize\implies{\sf{x=\dfrac{y_1+y_2}{2}}}$

where,

• y₁ = -1
• y₂ = y
• y = -1

Putting the values,

$\\ :\normalsize\implies{\sf{-1=\dfrac{-1+x}{2}}}$

$\\ :\normalsize\implies{\sf{-1\times2=-1+x}}$

$\\ :\normalsize\implies{\sf{-2=-1+x}}$

$\\ :\normalsize\implies{\sf{-2+1=x}}$

$\\ :\normalsize\implies{\sf{-1=x}}$

$\\ \normalsize\therefore\boxed{\bf{\pink{y_2=-1.}}}$

VerificatioN:-

$\\ :\normalsize\implies{\sf{(x,y)=\bigg\lgroup\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup}}$

where,

• x = 1
• y = -1
• x₁ = -3
• x₂ = 5
• y₁ = -1
• y₂ = -1

Putting the values,

$\\ :\normalsize\implies{\sf{(1,-1)=\bigg\lgroup\dfrac{-3+5}{2},\dfrac{-1+(-1)}{2}\bigg\rgroup}}$

$\\ :\normalsize\implies{\sf{(1,-1)=\bigg\lgroup\dfrac{-3+5}{2},\dfrac{-1-1}{2}\bigg\rgroup}}$

$\\ :\normalsize\implies{\sf{(1,-1)=\bigg\lgroup\dfrac{2}{2},\dfrac{-2}{2}\bigg\rgroup}}$

$\\ :\normalsize\implies{\sf{(1,-1)=\bigg\lgroup\cancel{\dfrac{2}{2}},\cancel{\dfrac{-2}{2}}\bigg\rgroup}}$

$\\ :\normalsize\implies{\sf{(1,-1)=(1,-1)}}$

$\\ \normalsize\therefore\boxed{\bf{LHS=RHS.}}$

• Hence verified.

The coordinates of point D is (5, -1) .

(for more reference refer to the attachment).