Question

find the length of the medians of a​

Answer 1

Given:

A triangle ABC.

A(0, -1)

B(2, 1)

C(0, 3)

To find:

Lengths of the medians of ABC.

Construction:

Plot the midpoints of AB, BC and CA.

And now join the vertex opposite to the midpoint, and we get the median.

Medians are namely AD, CE and BF.

Solution:

First, we'll find out the coordinates of these midpoints (D, E anf F) using the midpoint formula, then calculate their distance using the distance formula.

Midpoint of AB | Coordinates of E.

$\sf \Longrightarrow E(x, y) = \bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \bigg]$

Here:

x₁ = 0

x₂ = 2

y₁ = -1

y₂ = 1

$\sf \Longrightarrow E(x, y) = \bigg[\dfrac{0 + 2}{2} , \dfrac{-1 + 1}{2} \bigg]$

$\sf \Longrightarrow E(x, y) = \bigg[\dfrac{2}{2} , \dfrac{0}{2} \bigg]$

$\sf \Longrightarrow E(x, y) = \big[1 , 0\big]$

∴ The coordinates of E is (1, 0).

Midpoint of BC | Coordinates of D.

$\sf \Longrightarrow D(x, y) = \bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \bigg]$

Here:

x₁ = 2

x₂ = 0

y₁ = 1

y₂ = 3

$\sf \Longrightarrow D(x, y) = \bigg[\dfrac{2 + 0}{2} , \dfrac{1 + 3}{2} \bigg]$

$\sf \Longrightarrow D(x, y) = \bigg[\dfrac{2}{2} , \dfrac{4}{2} \bigg]$

$\sf \Longrightarrow D(x, y) = \big[1 , 2\big]$

∴ The coordinates of D are (1, 2).

Midpoint of AC | Coordinates of F.

$\sf \Longrightarrow F(x, y) = \bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \bigg]$

Here:

x₁ = 0

x₂ = 0

y₁ = -1

y₂ = 3

$\sf \Longrightarrow F(x, y) = \bigg[\dfrac{0 + 0}{2} , \dfrac{-1 + 3}{2} \bigg]$

$\sf \Longrightarrow F(x, y) = \bigg[\dfrac{0}{2} , \dfrac{2}{2} \bigg]$

$\sf \Longrightarrow F(x, y) = \big[0 , 1\big]$

∴ The coordinates of F are (0, 1).

Now, let's use the distance formula to find the lengths of the medians.

$\boxed{\sf Distance \ of \ any \ two \ given \ coordinates= \sqrt{\big(x_1 - x_2\big)^2 + \big(y_1 - y_2\big)^2}}$

Length of AD:

$\Longrightarrow \sf AD = \sqrt{\big(x_1 - x_2\big)^2 + \big(y_1 - y_2\big)^2}$

$\Longrightarrow \sf AD = \sqrt{\big(0 - 1\big)^2 + \big(-1 - 2\big)^2}$

$\Longrightarrow \sf AD = \sqrt{\big(- 1\big)^2 + \big(-3\big)^2}$

$\Longrightarrow \sf AD = \sqrt{1 + 9}$

$\Longrightarrow \sf AD = \sqrt{10} \ sq.units.$

Length of BF:

$\Longrightarrow \sf BF = \sqrt{\big(x_1 - x_2\big)^2 + \big(y_1 - y_2\big)^2}$

$\Longrightarrow \sf BF = \sqrt{\big(2 - 0\big)^2 + \big(1 - 1\big)^2}$

$\Longrightarrow \sf BF = \sqrt{\big(2\big)^2 + \big(0\big)^2}$

$\Longrightarrow \sf BF = 2 \ sq.units.$

Length of CE:

$\Longrightarrow \sf CE = \sqrt{\big(x_1 - x_2\big)^2 + \big(y_1 - y_2\big)^2}$

$\Longrightarrow \sf CE = \sqrt{\big(0 - 1\big)^2 + \big(3 - 0\big)^2}$

$\Longrightarrow \sf CE = \sqrt{\big(- 1\big)^2 + \big(3\big)^2}$

$\Longrightarrow \sf CE = \sqrt{1 + 9}$

$\Longrightarrow \sf CE = \sqrt{10} \ sq.units.$

Final Answers:

AD = √10 sq. units.

BF = 2 sq. units.

CE = √10 sq. units.

Answer 2

Answer:

Given:

A triangle ABC.

A(0, -1)

B(2, 1)

C(0, 3)

To find:

Lengths of the medians of ABC.

Construction:

Plot the midpoints of AB, BC and CA.

And now join the vertex opposite to the midpoint, and we get the median.

Medians are namely AD, CE and BF.

Solution:

First, we'll find out the coordinates of these midpoints (D, E anf F) using the midpoint formula, then calculate their distance using the distance formula.

Midpoint of AB | Coordinates of E.

$\sf \Longrightarrow E(x, y) = \bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \bigg]$

Here:

x₁ = 0

x₂ = 2

y₁ = -1

y₂ = 1

$\sf \Longrightarrow E(x, y) = \bigg[\dfrac{0 + 2}{2} , \dfrac{-1 + 1}{2} \bigg]$

$\sf \Longrightarrow E(x, y) = \bigg[\dfrac{2}{2} , \dfrac{0}{2} \bigg]$

$\sf \Longrightarrow E(x, y) = \big[1 , 0\big]$

∴ The coordinates of E is (1, 0).

Midpoint of BC | Coordinates of D.

$\sf \Longrightarrow D(x, y) = \bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \bigg]$

Here:

x₁ = 2

x₂ = 0

y₁ = 1

y₂ = 3

$\sf \Longrightarrow D(x, y) = \bigg[\dfrac{2 + 0}{2} , \dfrac{1 + 3}{2} \bigg]$

$\sf \Longrightarrow D(x, y) = \bigg[\dfrac{2}{2} , \dfrac{4}{2} \bigg]$

$\sf \Longrightarrow D(x, y) = \big[1 , 2\big]$

∴ The coordinates of D are (1, 2).

Midpoint of AC | Coordinates of F.

$\sf \Longrightarrow F(x, y) = \bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \bigg]$

Here:

x₁ = 0

x₂ = 0

y₁ = -1

y₂ = 3

$\sf \Longrightarrow F(x, y) = \bigg[\dfrac{0 + 0}{2} , \dfrac{-1 + 3}{2} \bigg]$

$\sf \Longrightarrow F(x, y) = \bigg[\dfrac{0}{2} , \dfrac{2}{2} \bigg]$

$\sf \Longrightarrow F(x, y) = \big[0 , 1\big]$

∴ The coordinates of F are (0, 1).

Now, let's use the distance formula to find the lengths of the medians.

$\boxed{\sf Distance \ of \ any \ two \ given \ coordinates= \sqrt{\big(x_1 - x_2\big)^2 + \big(y_1 - y_2\big)^2}}$

Length of AD:

$\Longrightarrow \sf AD = \sqrt{\big(x_1 - x_2\big)^2 + \big(y_1 - y_2\big)^2}$

$\Longrightarrow \sf AD = \sqrt{\big(0 - 1\big)^2 + \big(-1 - 2\big)^2}$

$\Longrightarrow \sf AD = \sqrt{\big(- 1\big)^2 + \big(-3\big)^2}$

$\Longrightarrow \sf AD = \sqrt{1 + 9}$

$\Longrightarrow \sf AD = \sqrt{10} \ sq.units.$

Length of BF:

$\Longrightarrow \sf BF = \sqrt{\big(x_1 - x_2\big)^2 + \big(y_1 - y_2\big)^2}$

$\Longrightarrow \sf BF = \sqrt{\big(2 - 0\big)^2 + \big(1 - 1\big)^2}$

$\Longrightarrow \sf BF = \sqrt{\big(2\big)^2 + \big(0\big)^2}$

$\Longrightarrow \sf BF = 2 \ sq.units.$

Length of CE:

$\Longrightarrow \sf CE = \sqrt{\big(x_1 - x_2\big)^2 + \big(y_1 - y_2\big)^2}$

$\Longrightarrow \sf CE = \sqrt{\big(0 - 1\big)^2 + \big(3 - 0\big)^2}$

$\Longrightarrow \sf CE = \sqrt{\big(- 1\big)^2 + \big(3\big)^2}$

$\Longrightarrow \sf CE = \sqrt{1 + 9}$

$\Longrightarrow \sf CE = \sqrt{10} \ sq.units.$

Final Answers:

AD = √10 sq. units.

BF = 2 sq. units.

CE = √10 sq. units.