Ex 4.3 Class 12 Maths Ncert Solutions​

Given vertices of the triangle are \( (1,0),(6,0),(4,3) \)
Let the vertices of the triangle be given by \((x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})\)
Area of triangle is given by \( \triangle=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| \)
Area of triangle \( =\triangle=\frac{1}{2}\left|\begin{array}{lll}1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1\end{array}\right| \)
Expanding the determinant along Row 1
\( \triangle=\frac{ 1 }{ 2 } \times[1 \times(0 \times 1-3 \times 1)-0 \times(6 \times 1-4 \times 1)+1 \times(6 \times 3-4 \) \( \times 0)] \)
\( \triangle=\frac{ 1 }{ 2 } \times[1 \times(0-3)-0+1 \times(18-0)] \)
\( \triangle=\frac{ 1 }{ 2 } \times(-3+18)=\frac{ 15 }{ 2 } \) sq. units
\( \therefore \triangle=\frac{ 15 }{ 2 } \) sq. units \( =7.5 \) sq. units

Given vertices of the triangle are \( (2,7),(1,1),(10,8) \)
Let the vertices of the triangle be given by \( \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) \)
Area of triangle is given by \( \triangle=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| \)
Area of triangle \( =\triangle=\frac{1}{2}\left|\begin{array}{ccc}2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1\end{array}\right| \)
Expanding the determinant along Row 1
\(\begin{array}{l}
\triangle=\frac{ 1 }{ 2 }\times[2 \times(1 \times 1-8 \times 1)-7 \times(1 \times 1-10 \times 1)+1 \times(8 \times 1- \\
1 \times 10)] \\
\triangle=\frac{ 1 }{ 2 } \times[2 \times(1-8)-7 \times(1-10)+1 \times(8-10)] \\
\triangle=\frac{ 1 }{ 2 } \times[2 \times(-7)-7 \times(-9)+1 \times(-2)]=\frac{ 1 }{ 2 } \times(-14+63-2) \text { sq.} \\
\text {units }
\end{array}\)
\( \triangle=\frac{1 }{ 2} \times 47 \) sq. units \( =\frac{ 47 }{ 2 } \) sq. units
\( \therefore \triangle=\frac{ 47 }{ 2 } \) sq. units \( =23.5 \) sq. units

Given vertices of the triangle are \( (-2,-3),(3,2),(-1,-8) \)
Let the vertices of the triangle be given by \( \left(x_{1}, x_{1}\right),\left(x_{2}, x_{2}\right),\left(x_{3}, x_{3}\right) \)
Area of triangle is given by \( \triangle=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| \)
Area of triangle \( =\triangle=\frac{1}{2}\left|\begin{array}{ccc}-2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1\end{array}\right| \)
Expanding the determinant along Row 1
\( \triangle=\frac{ 1 }{ 2 } \times \left|[(-2) \times(2 \times 1-(-8) \times 1)-(-3) \times(3 \times 1-(-1) \times 1)+1\right. \)
\(\left. \times(3 \times(-8)-2 \times(-1))] \right|\)
\( \triangle=\frac{ 1 }{ 2 } \times|[(-2) \times(2+8)+3 \times(3+1)+1 \times(-24+2)]| \)
\( \triangle=\frac{ 1 }{ 2 } \times|[(-2) \times 10+3 \times 4+1 \times(-22)]|=\frac{ 1 }{ 2 } \times \mid(-20+12- \)
\(22)|\) sq. units
\( \triangle=\frac{ 1 }{ 2 } \times|-30| \) sq. units \( =\frac{ 30 }{ 2 } \) sq. units
\( \therefore \triangle=\frac{ 30 }{ 2 } \) sq. units \( =15 \) sq. units

Given vertices of the triangle are
\( A(a, b+c), B(b, c+a), C(c, a+b) \)
Let the vertices of the triangle be given by \((x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})\)
Area of triangle is given by \( \triangle=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| \)
For points to be collinear area of triangle \( =\triangle=0 \)
So, we have to show that area of triangle formed by \( A B C \) is 0
Area of triangle \( =\triangle=\frac{1}{2}\left|\begin{array}{lll}a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1\end{array}\right| \)
Expanding the determinant along Row 1
\(
\triangle=\frac{ 1 }{ 2 }\times[\mathrm{a} \times\{(\mathrm{c}+\mathrm{a}) \times 1-(\mathrm{a}+\mathrm{b}) \times 1)-(\mathrm{b}+\mathrm{c}) \times\{\mathrm{b} \times 1-\mathrm{c} \times 1\}\)
\(
+1 \times\{\mathrm{b} \times(\mathrm{a}+\mathrm{b})-\mathrm{c} \times(\mathrm{c}+\mathrm{a})\}]\)
\(
\triangle=\frac{ 1 }{ 2 } \times[\mathrm{a} \times(\mathrm{c}+\mathrm{a}-\mathrm{a}-\mathrm{b})-(\mathrm{b}+\mathrm{c}) \times(\mathrm{b}-\mathrm{c})+1 \times(\mathrm{ab}+\mathrm{b}^{2}-\)
\(
\mathrm{ca})]\)
\(
\triangle=\frac{ 1 }{ 2 } \times\left[\mathrm{a} \times(\mathrm{c}-\mathrm{b})-\left(\mathrm{b}_{2}-\mathrm{c}_{2}\right)+1 \times\left(\mathrm{ab}+\mathrm{b}_{2}-\mathrm{c}_{2}-\mathrm{ca}\right)\right]\)
\(
\triangle=\frac{ 1 }{ 2 } \times\left(\mathrm{ac}-\mathrm{ab}-\mathrm{b}_{2}+\mathrm{c}_{2}+\mathrm{ab}+\mathrm{b}_{2}-\mathrm{c}_{2}-\mathrm{ca}\right) \text { sq units }\)
\(
\triangle=\frac{ 1 }{ 2 } \times 0 \text { sq units }\)
\(
\therefore \triangle=0
\)
\( \therefore \) Given vertices of the triangle are
\( \mathrm{A}(\mathrm{a}, \mathrm{b}+\mathrm{c}), \mathrm{B}(\mathrm{b}, \mathrm{c}+\mathrm{a}), \mathrm{C}(\mathrm{c}, \mathrm{a}+ \) b) are collinear

Given vertices of the triangle are \( (\mathrm{k}, 0),(4,0),(0,2) \)
Let the vertices of the triangle be given by \( \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) \)
Area of triangle is given by \( \triangle=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| \)
Given, Area of triangle \( =\triangle=4 \) sq. units
\(\begin{array}{l}
=\frac{1}{2} \left|\left| \begin{array}{ccc}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\right|=4 \end{array}\)
\(\Rightarrow 4=\frac{ 1 }{ 2 } |[\mathrm{k} \times(0 \times 1-2 \times 1)-0 \times(4 \times 1-0 \times 1)+1 \times(4 \times 2- \)\(0 \times 0)] |\)
\(\Rightarrow 4=\frac{ 1 }{ 2 } \times|[\mathrm{k} \times(0-2)-0+1 \times(8-0)]|\)
\(
\Rightarrow \pm 4 \times 2=-2 \mathrm{k}+8\)
\(
\Rightarrow 8=-2 \mathrm{k}+8 \text { and }-8=-2 \mathrm{k}+8\)
\(
\Rightarrow 8-8=-2 \mathrm{k} \text { and } 8+8=2 \mathrm{k}\)
\(
\Rightarrow 2 \mathrm{k}=0 \text { and } 16=2 \mathrm{k}\)
\(
\Rightarrow \mathrm{k}=0 \text { and } \mathrm{k}=8
\)

Given vertices of the triangle are \( (-2,0),(0,4),(0, k) \)
Let the vertices of the triangle be given by \((x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})\)
Area of triangle is given by \( \Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| \)
Given, Area of triangle \( =\Delta=4 \) sq. units
\(\begin{array}{l}
=\frac{1}{2}\left\|\begin{array}{ccc}
-2 & 0 & 1 \\
0 & 4 & 1 \\
0 & k & 1
\end{array}\right\|=4 \end{array}\)
\(\Rightarrow 4=\frac{1 }{ 2} |[(-2) \times(4 \times 1-\mathrm{k} \times 1)-0 \times(0 \times 1-0 \times 1)+1\)\( \times(0 \times \mathrm{k} -0 \times 4)] |\)
\(\Rightarrow 4=\frac{1 }{ 2} \times|[(-2) \times(4-\mathrm{k})-0+1 \times(0-0)]| \)
\(\Rightarrow 4 \times 2=|(-8+2 \mathrm{k})| \)
\(\Rightarrow \pm 8=2 \mathrm{k}-8 \)
\(\Rightarrow 8=2 \mathrm{k}-8 \text { and } \)
\(\Rightarrow-8=2 \mathrm{k}-8\)
\( \Rightarrow 8+8=2 \mathrm{k} \) and
\( \Rightarrow 8-8=2 \mathrm{k} \)
\( \Rightarrow \mathrm{k}=\frac{16 }{ 2} \) and
\( \Rightarrow \mathrm{k}=\frac{0 }{ 2} \)
\( \Rightarrow \mathrm{k}=8 \) and
\( \Rightarrow \mathrm{k}=0 \)

Equation of line joining points \( (x_{1}, \mathrm{y}_{1}) \ \& \ (x_{2}, y _{2}) \) is given by
\(\frac{1}{2}\left|\begin{array}{ccc}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x & y & 1
\end{array}\right|=0\)
Given points are \( (1,2) \) and \( (3,6) \)
Equation of line is given by \( \frac{1}{2}\left|\begin{array}{lll}1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1\end{array}\right|=0 \)
\(
\Rightarrow \frac{1 }{ 2} \times[1 \times(6 \times 1-y \times 1)-2 \times(3 \times 1-x \times 1)+1 \times(3 \times y-x \times\)
\(
6)]=0\)
\(
\Rightarrow[(6-y)-2 \times(3-x)+(3 y-6 x)]=0 \times 2\)
\(
\Rightarrow(6-y-6+2 x+3 y-6 x)=0\)
\(
\Rightarrow 2 y-4 x=0\)
\(
\Rightarrow y-2 x=0 \)
\(\Rightarrow y=2 x
\)
\( \therefore \) Required Equation of line is \( y=2 x \)

Equation of line joining points \( (x_{1}, y_{1}) ~\&~(x_{2}, y_{2}) \) is given by
\(\frac{1}{2}\left|\begin{array}{ccc}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x & y & 1
\end{array}\right|=0\)
Given points are \( (3,1) \) and \( (9,3) \)
Equation of line is given by \( \frac{1}{2}\left|\begin{array}{lll}3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1\end{array}\right|=0 \)
\(\Rightarrow \frac{1 }{ 2} \times[3 \times(3 \times 1-y\times 1)-1 \times(9 \times 1-x\times 1)+1 \times(9 \times y-x\times\)
\(
3)]=0\)
\(
\Rightarrow[3 \times(3-y)-1 \times(9-x)+(9 y-3 x)]=0 \times 2\)
\(
\Rightarrow(9-3 y-9+x+9 y-3 x)=0\)
\(
\Rightarrow 6 y-2 x=0\)
\(
\Rightarrow 2 x-6 y=0 \)
\(\Rightarrow x-3 y=0
\)
\( \therefore \) Required Equation of line is \( x-3 y=0 \)

Given vertices of the triangle are \( (2,-6),(5,4) \) and \( (k, 4) \).
Let the vertices of the triangle be given by \((x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) \)
Area of triangle is given by \( \triangle=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| \)
Given, Area of triangle \( =\triangle=35 \) sq. units
\(\begin{array}{l}
=\frac{1}{2}\left\|\begin{array}{ccc}
2 & -6 & 1 \\
5 & 4 & 1 \\
k & 4 & 1
\end{array}\right\|=35 \end{array}\)
\(\Rightarrow \pm 35=\frac{ 1 }{ 2 } \times[2 \times(4 \times 1-4 \times 1)-(-6) \times(5 \times 1-\mathrm{k} \times 1)+1 \times \)\((5 \times 4-\mathrm{k} \times 4)] \)
\(\Rightarrow \pm 35=\frac{ 1 }{ 2 } \times[2 \times(4-4)+6 \times(5-\mathrm{k})+1 \times(20-4 \mathrm{k})] \)
\(
\Rightarrow \pm 35 \times 2=(2 \times 0+30-6 \mathrm{k}+20-4 \mathrm{k}) \)
\(\Rightarrow \pm 70=30-6 \mathrm{k}+20-4 \mathrm{k} \)
\(\Rightarrow \pm 70=50-10 \mathrm{k} \)
\(\Rightarrow 70-50=-10 \mathrm{k} \text { and } \)
\(\Rightarrow-70-50=-10 \mathrm{k} \)
\(\Rightarrow 20=-10 \mathrm{k} \text { and }\)
\( \Rightarrow-120=-10 \mathrm{k} \)
\(\Rightarrow \mathrm{k}=-\frac{ 20 }{ 10 } \text { and } \)
\(\Rightarrow \mathrm{k}=\frac{ 120 }{ 10 } \)
\(
\Rightarrow \mathrm{k}=-2 \text { and } \)
\(\Rightarrow \mathrm{k}=12 \)