Question

If the vertices of a triangle ∆ABC are (–5, 2), (+9, –3) and (–3, +5), then find the area of the

triangle formed by the mid points of sides of ∆ABC .​

Answer 1

The area of  the triangle formed by the mid-points of sides of ∆ABC is 3 sq units.

Step-by-step explanation:

We are given that the vertices of a triangle ∆ABC are (-5, 2), (9, -3) and (-3, 5).

Let the vertices of ∆ABC be denoted by A(-5, 2), B(9, -3) and C(-3, 5). Also, let P = mid-point of side AB, Q = mid-point of side AC and R = mid-point of side BC.

Now, the mid-point formula states that;

If there are two points ($x_1,y_1$) and ($x_2,y_2$), then the mid-point of these two coordinates is given by = $(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2})$

So, mid-point of side AB with A(-5, 2) and B(9, -3) is given by;

            =  $(\frac{-5+9}{2} ,\frac{2+(-3)}{2})$ = $(\frac{4}{2} ,\frac{1}{2})$

            =  $(2 ,0.5)$

Similarly, mid-point of side AC with A(-5, 2) and C(-3, 5) is given by;

            =  $(\frac{-5+(-3)}{2} ,\frac{2+5}{2})$ = $(\frac{-8}{2} ,\frac{7}{2})$

            =  $(-4,3.5)$

And mid-point of side BC with B(9, -3) and C(-3, 5) is given by;

            =  $(\frac{9+(-3)}{2} ,\frac{-3+5}{2})$ = $(\frac{6}{2} ,\frac{2}{2})$

            =  $(3 ,1)$

So, the vertices of ∆PQR are calculated as P(2, 0.5), Q(-4, 3.5) and R(3,1).

Now, the area of the traingle having vertices ($x_1,y_1$), ($x_2,y_2$) and ($x_3,y_3$) is given by;

Area of triangle =  $|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2} |$

Substituting the value of P, Q and R coordinates values we get;

                     =  $|\frac{2(3.5-1)+(-4)(1-0.5)+3(0.5-3.5)}{2} |$

                     =  $| \frac{2(2.5)+(-4)(0.5)+3(-3)}{2} |$

                     =  $| \frac{5-2-9}{2} |$

                     =  $| \frac{-6}{2} |$  = 3 sq. units

Hence, the area of the  triangle formed by the mid-points of sides of ∆ABC is 3 sq. units.