Chapter 4: Determinants

Ans: Given vertices,\[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\]

Thus, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}   \text{1} & \text{0} & \text{1}  \\   \text{6} & \text{0} & \text{1}  \\   \text{4} & \text{3} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta =\dfrac{\text{1}}{\text{2}}\left[ \text{1}\left( \text{0-3} \right)\text{-0}\left( \text{6-4} \right)\text{+1}\left( \text{18-0} \right) \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-3+18} \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{15}}{\text{2}}\] square units

$\therefore $Area of the triangle with vertices \[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\] is \[\dfrac{\text{15}}{\text{2}}\] square units.

Ans: Given vertices,\[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{2} & \text{7} & \text{1}  \\  \text{1} & \text{1} & \text{1}  \\   \text{10} & \text{8} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{1-8} \right)\text{-7}\left( \text{1-10} \right)\text{+1}\left( \text{8-10} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{-7} \right)\text{-7}\left( \text{-9} \right)\text{+1}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-16+63} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{47}}{\text{2}}\] square units

$\therefore $Area of the triangle with vertices \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\] is \[\dfrac{47}{\text{2}}\] square units.

Ans: Given vertices,\[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\]

The area of the triangle with vertices \[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\] is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{-2} & \text{-3} & \text{1}  \\  \text{3} & \text{2} & \text{1}  \\  \text{-1} & \text{-8} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-2}\left( \text{2+8} \right)\text{+3}\left( \text{3+1} \right)\text{+1}\left( \text{-24+2} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-20+12-22} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ }=-\dfrac{\text{30}}{\text{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-15}\]

$\therefore $The area of the triangle with vertices \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]is \[\left| \text{-15} \right|\text{=15}\] square units.

Ans: To show that the points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{, C}\left( \text{c, a +b} \right)\] are collinear, the area of the triangle formed by these points as vertices should be zero.

$\therefore $ Area of \[\text{ }\!\!\Delta\!\!\text{ ABC}\] is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}   \text{a} & \text{b+c} & \text{1}  \\   \text{b} & \text{c+a} & \text{1}  \\   \text{c} & \text{a+b} & \text{1}  \\ \end{matrix} \right|\]

Applying the row operations, \[R_2\to R_2-R_1\] and \[R_3\to R_3-R_1\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}  \text{a} & \text{b+c} & \text{1}  \\ \text{b-a} & \text{a-b} & \text{0}  \\ \text{c-a} & \text{a-c} & \text{0}  \\ \end{matrix} \right|\]

Taking out $\left( a-b \right)$ and $\left( c-a \right)$ common from \[R_2\] and \[R_3\] respectively,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left( \text{a-b} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\ \text{-1} & \text{1} & \text{0}  \\ \text{1} & \text{-1} & \text{0}  \\ \end{matrix} \right|\]

Applying the row operation \[R_3\to R_3\text{+}R_2\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left( \text{a-b} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\   \text{-1} & \text{1} & \text{0}  \\  \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right|\]

We know that when all the elements of a row or a column in a determinant are zero then the value of the determinant is always zero.

\[\therefore \Delta \text{=0}\]                                           

Thus, the area of the triangle formed by points \[\text{A}\] , \[\text{B}\] and \[\text{C}\] is zero.

Hence, the points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)\] are collinear.

Ans: Given vertices are \[\left( \text{k,0} \right)\text{,}\left( \text{4,0} \right)\text{,}\left( \text{0,2} \right)\].

We know, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}  \text{k} & \text{0} & \text{1}  \\  \text{4} & \text{0} & \text{1}  \\  \text{0} & \text{2} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{k}\left( \text{0-2} \right)\text{-0}\left( \text{4-0} \right)\text{+1}\left( \text{8-0} \right) \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-2k+8} \right]\]

\[\therefore \Delta =-k+4\]

Since the area is given to be \[\text{4}\] square units, thus

$-k+4=\pm 4$

When \[-k+4=-4\]

\[\therefore k=8\].

When \[-k+4=4\]

\[\therefore k=0\].

Hence, \[\text{k=0,8}\].

Ans: Given vertices are \[\text{(-2,0),}\left( \text{0,4} \right)\text{,}\left( \text{0,k} \right)\].

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}  \text{-2} & \text{0} & \text{1}  \\  \text{0} & \text{4} & \text{1}  \\  \text{0} & \text{k} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ -2\left( 4-k \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ }=k-4\]

Since the area is given to be \[\text{4}\] square units, thus

\[k-4=\pm 4\]

When \[k-4=-4\]

\[\therefore k=0\].

When \[k-4=4\]

\[\therefore k=8\].

Hence, \[k=0,8\].

Ans: Let us assume a point, \[\text{P}\left( \text{x, y} \right)\] on the line joining points \[\text{A}\left( \text{1,2} \right)\] and \[\text{B}\left( \text{3,6} \right)\] .

Then, the points \[\text{A}\],$B$ and $P$ are collinear.

Thus, the area of the triangle \[\text{ABP}\] will be zero.

\[\therefore \dfrac{\text{1}}{\text{2}}\left| \begin{matrix}  \text{1} & \text{2} & \text{1}  \\ \text{3} & \text{6} & \text{1}  \\  \text{x} & \text{y} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow \dfrac{\text{1}}{\text{2}}\left[ \text{1}\left( \text{6-y} \right)\text{-2}\left( \text{3-x} \right)\text{+1}\left( \text{3y-6x} \right) \right]\text{=0}\]

\[\Rightarrow \text{6-y-6+2x+3y-6x=0}\]

\[\Rightarrow \text{2y-4x=0}\]

\[\Rightarrow \text{y=2x}\]

$\therefore $ The equation of the line joining the given points is \[y=2x\].

Ans: Let us assume a point, \[\text{P}\left( \text{x, y} \right)\] on the line joining points \[\text{A}\left( \text{3,1} \right)\] and \[\text{B}\left( \text{9,3} \right)\]. 

Then, the points \[\text{A}\],$B$ and $P$ are collinear.

Thus, the area of the triangle \[\text{ABP}\] will be zero.

\[\therefore \dfrac{\text{1}}{\text{2}}\left| \begin{matrix}   \text{3} & \text{1} & \text{1}  \\   \text{9} & \text{3} & \text{1}  \\   \text{x} & \text{y} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow \dfrac{\text{1}}{\text{2}}\left[ \text{3}\left( \text{3-y} \right)\text{-1}\left( \text{9-x} \right)\text{+1}\left( \text{9y-3x} \right) \right]\text{=0}\]

\[\Rightarrow \text{9-3y-9+x+9y-3x=0}\]

\[\Rightarrow \text{6y-2x=0}\]

\[\Rightarrow \text{x-3y=0}\]

$\therefore $ The equation of the line joining the given points is \[\text{x-3y=0}\] .

Ans: Given vertices, \[\text{(2,-6)}\],\[\left( \text{5,4} \right)\] and \[\left( \text{k,4} \right)\]

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix}   \text{2} & \text{-6} & \text{1}  \\   \text{5} & \text{4} & \text{1}  \\   \text{k} & \text{4} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{4-4} \right)\text{+6}\left( \text{5-k} \right)\text{+1}\left( \text{20-4k} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{50-10k} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =25-5k}\]

Given, the area of the triangle is \[\text{35}\] square units.

Thus, we have:

\[\Rightarrow 25-5k=\pm 35\]

\[\Rightarrow 5\left( 5-k \right)=\pm 35\]

\[\Rightarrow 5-k=\pm 7\].

When \[5-k=7\]

\[\therefore k=-2\].

When \[5-k=-7\]

\[\therefore k=12\].

Hence, \[k=12,-2\] .

Thus, D. \[12,-2\] is the correct option.