Exercise-4.2, Class 12th, Maths, Chapter 4, NCERT

Question 1 (Area of triangles)

$$\text{Find area of the triangle whose vertices are:}$$
$$\text{(i) }(1,0),(6,0),(4,3)\text{(ii) }(2,7),(1,1),(10,8)$$

$$\text{(iii) }(-2,-3),(3,2),(-1,-8)\mathrm{.}$$

Method. Area $={\displaystyle \frac{1}{2}}\mid x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\mid$

(i) $x_1=1,y_1=0;x_2=6,y_2=0;x_3=4,y_3=3$

$$S=\frac{1}{2}\mid 1(0-3)+6(3-0)+4(0-0)\mid$$

$$=\frac{1}{2}\mid -3+18+0\mid =\frac{1}{2}(15)=\boxed{{\displaystyle \frac{15}{2}}}\mathrm{.}$$

(ii) $(2,7),(1,1),(10,8)$

$$S=\frac{1}{2}\mid 2(1-8)+1(8-7)+10(7-1)\mid$$

$$=\frac{1}{2}\mid -14+1+60\mid =\frac{1}{2}(47)=\boxed{{\displaystyle \frac{47}{2}}}\mathrm{}$$

(iii) $(-2,-3),(3,2),(-1,-8)$

$$S=\frac{1}{2}\mid -2(2-(-8))+3((-8)-(-3))+(-1)((-3)-2)\mid$$

$$=\frac{1}{2}\mid -20-15+5\mid =\frac{1}{2}(30)=\boxed{{\displaystyle 15}}\mathrm{}$$

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Question 2 (Collinearity)

$$\text{Show that }A(a,b+c),B(b,c+a),C(c,a+b)\text{ are collinear.}$$

Solution. Points are collinear iff area of triangle $ABC$ is zero. Compute

$$\mathrm{\Delta }=\mid \begin{array}{ccc}a& b+c& 1\\ b& c+a& 1\\ c& a+b& 1\end{array}\mid =a((c+a)-(a+b))+b((a+b)-(b+c))+c((b+c)-(c+a))\mathrm{}$$

Simplify each bracket:

$$a(c-b)+b(a-c)+c(b-a)=ac-ab+ab-bc+bc-ac=0$$

Hence $\mathrm{\Delta }=0$ so $A,B,C$ are collinear. $\boxed{{\displaystyle }}$​

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Question 3 (Values of $k$ for area = 4)

$$\text{Find }k\text{ if area }=4\text{ for:}$$
$$\text{(i) }(k,0),(4,0),(0,2)\text{(ii) }(-2,0),(0,4),(0,k)\mathrm{.}$$

(i) Using determinant formula:

$$\mathrm{\Delta }=k(0-2)+4(2-0)+0(0-0)=-2k+8.$$

Area $=\frac{1}{2}\mathrm{\mid }\mathrm{\Delta }\mathrm{\mid }=4\Rightarrow \mathrm{\mid }\text{\hspace{0.17em}⁣}-2k+8\text{\hspace{0.17em}⁣}\mathrm{\mid }=8$
So $-2k+8=8$ or $-2k+8=-8$ giving $k=0$ or $k=8$
$\boxed{{\displaystyle k=0\text{ or }8}}$

(ii)

$$\mathrm{\Delta }=-2(4-k)+0+0=-8+2k\mathrm{}$$

$\frac{1}{2}\mathrm{\mid }-8+2k\mathrm{\mid }=4\Rightarrow \mathrm{\mid }-8+2k\mathrm{\mid }=8$. So $-8+2k=8$ or $-8+2k=-8$
gives $k=8$ or $k=0$
$\boxed{{\displaystyle k=0\text{ or }8.}}$

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Question 4 (Equation of a line using determinants)

$$\text{(i) line joining }(1,2)\text{ and }(3,6)\mathrm{.}\text{(ii) line joining }(3,1)\text{ and }(9,3)\mathrm{.}$$

Use determinant form for line through $(x_1,y_1),(x_2,y_2)$

$$\mid \begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\mid =0.$$

(i) $\mid \begin{array}{ccc}x& y& 1\\ 1& 2& 1\\ 3& 6& 1\end{array}\mid =0$

$$\Rightarrow x(2-6)-y(1-3)+1(6-6)=-4x+2y=0\Rightarrow \boxed{{\displaystyle y=2x}}\mathrm{.}$$

(ii) $\mid \begin{array}{ccc}x& y& 1\\ 3& 1& 1\\ 9& 3& 1\end{array}\mid =0$

$$\Rightarrow x(1-3)-y(3-9)+1(9-3)=-2x+6y=0\Rightarrow \boxed{{\displaystyle x-3y=0}}\mathrm{.}$$

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Question 5 (Find $k$ from given area = 35)

$$\text{If area of triangle is }35\text{ with vertices }(2,-6),(5,4),(k,4)\mathrm{}$$

The points $(5,4)$ and $(k,4)$ have same $y$-coordinate, so base $=\mathrm{\mid }k-5\mathrm{\mid }$, height from $(2,-6)$ to line $y=4$ is $4-(-6)=10$

Area $=\frac{1}{2}\cdot \text{base}\cdot \text{height}=\frac{1}{2}\cdot \mathrm{\mid }k-5\mathrm{\mid }\cdot 10=5\mathrm{\mid }k-5\mathrm{\mid }=35$
So $\mathrm{\mid }k-5\mathrm{\mid }=7\Rightarrow k-5=\pm 7\Rightarrow k=12\text{ or }k=-2.$

$\boxed{{\displaystyle k=12\text{ or }k=-2}}$ (So the correct choice from the options is the pair $12,-2$)