The co-ordinates of three points A, B, C are (3, 4) (–4, 3) and (8, –6) respectively. Let us find the area of triangle and the perpendicular length drawn from the point A on BC.

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

A = (x₁, y₁) = (3, 4)

B = (x₂, y₂) = (-4, 3)

C = (x₃, y₃) = (8, -6)

Substituting values

⇒ Area = × [3(3 – (-6)) + (-4)(-6 – 4) + 8(4 – 3)]

⇒ Area = × [27 + 40 + 8]

⇒ Area = × 75

⇒ Area = 37.5 unit² … (i)

Consider AD is the perpendicular dropped on BC as shown

So, for ΔABC BC is the base then AD is the height

Area of triangle = × base × height … (ii)

Length of BC can be calculated by distance formula because coordinates of B and C are known

⇒ Distance BC =

=

=

= √225

= 15 units

Distance BC = 15 units

Using (i) and (ii)

⇒ 37.5 = × BC × AD

⇒ 37.5 = × 15 × AD

⇒ 75 = 15 × AD

⇒ AD = 3 units

area of triangle is 37.5 unit² and the perpendicular length drawn from the point A on BC i.e. AD = 3 units