Question

Find the midpoint of the line segment joining the points (-2, 6) and (3, 1).​

Answer 1

Given :

• A(-2, 6)
• B(3, 1)

To Find :

The mid-point of the line segment.

Solution :

Analysis :

Here the concept of mid-point formula is used. We have to use the mid point formula and substitute the values.

Required Formula :

The required mid point formula is,

$\\ \boxed{\bf(x,y)=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)}$

where,

• (x₁, y₁) = Coordinates of first point
• (x₂, y₂) = Coordinates of Second point

Explanation :

• A(-2, 6)
• B(3, 1)

Using the formula,

$\\ \bf(x,y)=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

where,

• x₁ = -2
• x₂ = 3
• y₁ = 6
• y₂ = 1

Substituting the required values,

$\\ \sf\implies(x,y)=\left(\dfrac{(-2)+3}{2},\dfrac{6+1}{2}\right)$

$\\ \sf\implies(x,y)=\left(\dfrac{-2+3}{2},\dfrac{6+1}{2}\right)$

$\\ \sf\implies(x,y)=\left(\dfrac{1}{2},\dfrac{7}{2}\right)$

$\\ \sf\implies(x,y)=\left(\dfrac{1}{2},3\dfrac{1}{2}\right)$

$\\ \therefore\boxed{\bf(x,y)=\left(\dfrac{1}{2},3\dfrac{1}{2}\right)}$

$\therefore\underline{\sf The\ mid-point\ is\ \bf{\left(\dfrac{1}{2},3\dfrac{1}{2}\right)}}.$

Explore More :

• Section-Formula :

$\\ \bf(x,y)=\Bigg\lgroup{\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\Bigg\rgroup}$

• Centroid of a Triangle :

$\\ \bf(x,y)=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$