Question

Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from B to AC.

Answer 1

The area of a triangle is =54 square units

The perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

$x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and y_3=8$

The area of a triangle is= $\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]$

=$|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|$

=$|-54|$ = 54 square units

The length of AC = $\sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}$

                          = $\sqrt{(2-12)^{2} +(1-8)^2}$

                         =$\sqrt{149}$ units

Let the perpendicular distance from B to AC be = x

According To Problem

$\frac{1}{2} \times x \times \sqrt{149} = 54$

⇔$x =\frac{108}{\sqrt{149} }$ units

Therefore the perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$