Chapter 7, Problem 129CR

To determine

The area, if possible, of the triangle, the coordinates of whose vertices are given or show that the points are collinear.

Answer to Problem 129CR

The area of the triangle is equal to $\frac{13}{2}$ sq. units.

Explanation of Solution

Given: The given coordinates of the points are $(2,4)$ , $(5,6)$ and $(4,1)$ .

Concept Used: Consider a triangle whose vertices have the coordinates $(x_1,y_1)$ , $(x_2,y_2)$ and $(x_3,y_3)$ . Then, area of this triangle is given by-

  $\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$ , where the modulus value is taken if the determinant has a negative value.

In case the three points are collinear, the value of the determinant will be zero.

Calculations: The given coordinates are $(2,4)$ , $(5,6)$ and $(4,1)$ . Therefore, area of the triangle having these points as its vertices is-

  $\Delta =\frac{1}{2}\left|\begin{array}{ccc}2& 4& 1\\ 5& 6& 1\\ 4& 1& 1\end{array}\right|$

  $=\frac{1}{2}\left\{2\left|\begin{array}{cc}6& 1\\ 1& 1\end{array}\right|-4\left|\begin{array}{cc}5& 1\\ 4& 1\end{array}\right|+1\left|\begin{array}{cc}5& 6\\ 4& 1\end{array}\right|\right\}$ (Expanding along $R_1$ )

  $=\frac{1}{2}\left\{2(6-1)-4(5-4)+1(5-24)\right\}$

  $=\frac{1}{2}\left\{2\times 5-4\times 1+1\times (-19)\right\}$

  $=\frac{1}{2}\left\{10-4-19\right\}$

  $=\frac{-13}{2}$

Therefore, area of the triangle is $\frac{13}{2}$ sq. units.