Question

Find the lengths of medians of a triangle whose vertices are (3, −6), (7, 4), and (−5, 2).

Answer 1

Formula:

I. Midpoint of points (x₁, y₁) and (x₂, y₂):

`((x_1 + x_2) / 2, (y_1 + y_2) / 2)`

II. Distance between (x₁, y₁) and (x₂, y₂):

`sqrt[(x_2 − x_1)^2 + (y_2 − y_1)^2]`

III. Median-length formula (from side lengths a, b, c):

Median to side a:

`m_a = (1/2) * sqrt(2b^2 + 2c^2 − a^2)`  ....(Similarly cyclic for m_b, m_c)

Given vertices:

A(3, −6), B(7, 4), C(−5, 2)

⇒ Median from A - find the midpoint of BC:

Midpoint M_a = `((7 + (−5))/2, (4 + 2)/2)`

∴ Midpoint M_a = (1, 3)

Distance AM_a:

AM_a = `sqrt[(3 − 1)^2 + (−6 − 3)^2]`

AM_a = `sqrt[2^2 + (−9)^2]`

AM_a = `sqrt[4 + 81]`

∴ AM_a = `sqrt(85)`

⇒ Median from B - midpoint of AC:

Midpoint M_b = `((3 + (−5))/2, (−6 + 2)/2)`

∴ Midpoint M_b = (−1, −2)

Distance BM_b:

BM_b = `sqrt[(7 − (−1))^2 + (4 − (−2))^2]`

BM_b = `sqrt[8^2 + 6^2]`

BM_b = `sqrt[64 + 36]`

BM_b = `sqrt(100)`

∴ BM_b = 10

⇒ Median from C - midpoint of AB:

Midpoint M_c = `((3 + 7)/2, (−6 + 4)/2)`

∴ Midpoint M_c = (5, −1)

Distance CM_c:

CM_c = `sqrt[(−5 − 5)^2 + (2 − (−1))^2]`

CM_c = `sqrt[(−10)^2 + 3^2]`

CM_c = `sqrt[100 + 9]`

∴ CM_c = `sqrt(109)`

Here, let’s apply the median-length formula:

Computing side squares:

a = BC = a² = 148;

b = CA = b² = 128;

c = AB = c² = 116

⇒ m_a = `(1/2) * sqrt(2b^2 + 2c^2 − a^2)`

m_a = `(1/2) * sqrt(256 + 232 − 148)`

m_a = `(1/2) * sqrt(340)`

∴ m_a = `sqrt(85)`

⇒ m_b = `(1/2) * sqrt(2c^2 + 2a^2 − b^2)`

m_b = `(1/2) * sqrt(232 + 296 − 128)`

m_b = `(1/2) * sqrt(400)`

∴ m_b = 10

⇒ m_c = `(1/2) * sqrt(2a^2 + 2b^2 − c^2)`

m_c = `(1/2) * sqrt(296 + 256 − 116)`

m_c = `(1/2) * sqrt(436)`

∴ m_c = `sqrt(109)`

Hence, final lengths of the medians:

• Median from A = `sqrt(85) = (≈ 9.2195)`
• Median from B = 10
• Median from C = `sqrt(109) = (≈ 10.4403)`