The vertices of ∆ ABC are ( - 2, 1), (5, 4) and (2, - 3) respectively. Find the area of the triangle and the length of the altitude through A.

Given: vertices of ∆ ABC are ( - 2, 1), (5, 4) and (2, - 3).

To find: Area of ∆ ABC.

Length of altitude.

Formula Used:

Area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃)
= 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

The distance between the points (x₁,y₁) and (x₂,y₂) is:

Distance

Explanation:

Let three vertices be A (−2, 1) and B (5, 4) and C(2, −3)

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)
= 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

Area of ∆ABC

= 1/2 | - 2(4 – ( - 3)) + 5( - 3 - 1) + 2(1 – 4)|

= 1/2 | - 14 - 20 - 6|

= 20 sq. units

Now to find the length of BC,

By distance formula,

XY =

For BC,

BC =

= √58 sq. units

Area of ∆ABC = 1/2 × Base × Altitude

∴ 20 = 1/2 × √58 × Altitude

∴ Altitude = units

Hence, the length of altitude through A is units.