Class 12 Maths NCERT Solutions Chapter 4 Ex 4.2 – Area of a Triangle

The area of a triangle with vertices (x_1,y_1),\ (x_2,y_2),\ (x_3,y_3) is given by

\Delta = \dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

(i) For (1,0),\ (6,0),\ (4,3):

\Delta = \dfrac{1}{2}\left|1(0-3)+6(3-0)+4(0-0)\right| = \dfrac{1}{2}\left|-3+18+0\right| = \dfrac{1}{2}(15)

(ii) For (2,7),\ (1,1),\ (10,8):

\Delta = \dfrac{1}{2}\left|2(1-8)+1(8-7)+10(7-1)\right| = \dfrac{1}{2}\left|-14+1+60\right| = \dfrac{1}{2}(47)

Three points are collinear exactly when the area of the triangle they form is zero. Using the area formula:

\Delta = \dfrac{1}{2}\left|a\big[(c+a)-(a+b)\big]+b\big[(a+b)-(b+c)\big]+c\big[(b+c)-(c+a)\big]\right|

Simplifying each bracket: (c+a)-(a+b) = c-b,   (a+b)-(b+c) = a-c,   (b+c)-(c+a) = b-a

So \Delta = \dfrac{1}{2}\left|a(c-b)+b(a-c)+c(b-a)\right| = \dfrac{1}{2}\left|ac-ab+ab-bc+bc-ac\right| = \dfrac{1}{2}|0|

(i) For (k,0),\ (4,0),\ (0,2):

\Delta = \dfrac{1}{2}\left|k(0-2)+4(2-0)+0(0-0)\right| = \dfrac{1}{2}\left|-2k+8\right| = |4-k|

Setting \Delta = 4: |4-k| = 4 \Rightarrow 4-k = 4 \text{ or } 4-k = -4

(ii) For (-2,0),\ (0,4),\ (0,k):

\Delta = \dfrac{1}{2}\left|-2(4-k)+0(k-0)+0(0-4)\right| = \dfrac{1}{2}\left|-8+2k\right| = |k-4|

Setting \Delta = 4: |k-4| = 4 \Rightarrow k-4 = 4 \text{ or } k-4 = -4

Let (x,y) be any point on the line through (1,2) and (3,6). Then the points (x,y),\ (1,2),\ (3,6) are collinear, so the area they enclose is zero:

\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0

Expanding along R_1:

x(2-6)-y(1-3)+1(6-6) = 0

-4x+2y+0 = 0 \Rightarrow 2y = 4x

\Delta = \dfrac{1}{2}\left|2(4-4)+5\big(4-(-6)\big)+k(-6-4)\right| = \dfrac{1}{2}\left|0+50-10k\right| = |25-5k|

Setting \Delta = 35: |25-5k| = 35 \Rightarrow 25-5k = 35 \text{ or } 25-5k = -35

First case: -5k = 10 \Rightarrow k = -2. Second case: -5k = -60 \Rightarrow k = 12.