MATHEMATICS CLASS- 12
CHAPTER-4
(DETERMINANTS)
Exercise 4.2
Question 1
Find the area of the triangle with vertices at the points given in each of the following.
(i)
The vertices of the triangle are
$$ (1,0),\ (6,0),\ (4,3). $$Solution:
Using the formula for the area of a triangle,
$$ \text{Area} = \frac12 \left| \begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix} \right|. $$Substituting the given coordinates,
$$ \text{Area} = \frac12 \left| \begin{vmatrix} 1&0&1\\ 6&0&1\\ 4&3&1 \end{vmatrix} \right|. $$Expanding along the first row,
$$ \begin{aligned} \text{Area} &= \frac12 \left| 1 \begin{vmatrix} 0&1\\ 3&1 \end{vmatrix} +0 + 1 \begin{vmatrix} 6&0\\ 4&3 \end{vmatrix} \right| \\[2mm] &= \frac12 \left| 1(0-3)+18 \right| \\[2mm] &= \frac12|15| \\[2mm] &= \frac{15}{2}. \end{aligned} $$Hence, the area of the triangle is
$$ \boxed{\frac{15}{2}\text{ square units}} $$(ii)
The vertices of the triangle are
$$ (2,7),\ (1,1),\ (10,8). $$Solution:
Using the formula for the area of a triangle,
$$ \text{Area} = \frac12 \left| \begin{vmatrix} 2&7&1\\ 1&1&1\\ 10&8&1 \end{vmatrix} \right|. $$Expanding along the first row,
$$ \begin{aligned} \text{Area} &= \frac12 \left| 2 \begin{vmatrix} 1&1\\ 8&1 \end{vmatrix} - 7 \begin{vmatrix} 1&1\\ 10&1 \end{vmatrix} + 1 \begin{vmatrix} 1&1\\ 10&8 \end{vmatrix} \right| \\[2mm] &= \frac12 \left| 2(1-8)-7(1-10)+(8-10) \right| \\[2mm] &= \frac12 \left| -14+63-2 \right| \\[2mm] &= \frac12(47) \\[2mm] &= \frac{47}{2}. \end{aligned} $$Hence, the area of the triangle is
$$ \boxed{\frac{47}{2}\text{ square units}} $$(iii)
The vertices of the triangle are
$$ (-2,-3),\ (3,2),\ (-1,-8). $$Solution:
Using the formula for the area of a triangle,
$$ \text{Area} = \frac12 \left| \begin{vmatrix} -2&-3&1\\ 3&2&1\\ -1&-8&1 \end{vmatrix} \right|. $$Expanding along the first row,
$$ \begin{aligned} \text{Area} &= \frac12 \left| -2 \begin{vmatrix} 2&1\\ -8&1 \end{vmatrix} +3 \begin{vmatrix} 3&1\\ -1&1 \end{vmatrix} + 1 \begin{vmatrix} 3&2\\ -1&-8 \end{vmatrix} \right| \\[2mm] &= \frac12 \left| -2(2+8)+3(3+1)+(-24+2) \right| \\[2mm] &= \frac12 \left| -20+12-22 \right| \\[2mm] &= \frac12(30) \\[2mm] &= 15. \end{aligned} $$Hence, the area of the triangle is
$$ \boxed{15\text{ square units}} $$Question 2
Show that the points
$$ A(a,\;b+c),\quad B(b,\;c+a),\quad C(c,\;a+b) $$are collinear.
Solution:
To show that the given points are collinear, we find the area of the triangle formed by them.
If the area of the triangle is zero, then the three points are collinear.
Using the determinant formula,
$$ \text{Area} = \frac12 \left| \begin{vmatrix} a&b+c&1\\ b&c+a&1\\ c&a+b&1 \end{vmatrix} \right|. $$Expand the determinant along the first row.
$$ \begin{aligned} \text{Area} &= \frac12 \left| a \begin{vmatrix} c+a&1\\ a+b&1 \end{vmatrix} -(b+c) \begin{vmatrix} b&1\\ c&1 \end{vmatrix} + \begin{vmatrix} b&c+a\\ c&a+b \end{vmatrix} \right|. \end{aligned} $$Now evaluate each minor.
$$ \begin{aligned} \begin{vmatrix} c+a&1\\ a+b&1 \end{vmatrix} &=(c+a)-(a+b) \\[2mm] &=c-b. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} b&1\\ c&1 \end{vmatrix} &=b-c. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} b&c+a\\ c&a+b \end{vmatrix} &=b(a+b)-c(c+a) \\[2mm] &=ab+b^2-c^2-ac. \end{aligned} $$Substituting these values,
$$ \begin{aligned} \text{Area} &= \frac12 \left| a(c-b) -(b+c)(b-c) +ab+b^2-ac-c^2 \right|. \end{aligned} $$Since
$$ (b+c)(b-c)=b^2-c^2, $$we get
$$ \begin{aligned} \text{Area} &= \frac12 \left| ac-ab-b^2+c^2+ab+b^2-ac-c^2 \right| \\[2mm] &= \frac12|0| \\[2mm] &=0. \end{aligned} $$Since the area of the triangle is zero, the given points are collinear.
Hence proved.
Question 3
Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are
(i)
$$ (k,0),\ (4,0),\ (0,2) $$Solution:
Using the formula for the area of a triangle,
$$ \text{Area} = \frac12 \left| \begin{vmatrix} k&0&1\\ 4&0&1\\ 0&2&1 \end{vmatrix} \right|. $$Since the area is $4$ square units,
$$ \frac12 \left| \begin{vmatrix} k&0&1\\ 4&0&1\\ 0&2&1 \end{vmatrix} \right| =4. $$Expanding the determinant along the first row,
$$ \begin{aligned} \begin{vmatrix} k&0&1\\ 4&0&1\\ 0&2&1 \end{vmatrix} &= k \begin{vmatrix} 0&1\\ 2&1 \end{vmatrix} +0 + 1 \begin{vmatrix} 4&0\\ 0&2 \end{vmatrix} \\[2mm] &= k(0-2)+8 \\[2mm] &= 8-2k. \end{aligned} $$Therefore,
$$ \frac12|8-2k|=4. $$ $$ |8-2k|=8. $$Hence,
Case 1:
$$ 8-2k=8 $$ $$ k=0. $$Case 2:
$$ 8-2k=-8 $$ $$ -2k=-16 $$ $$ k=8. $$Hence, the required values of $k$ are
$$ \boxed{k=0\ \text{or}\ k=8.} $$(ii)
The vertices of the triangle are
$$ (-2,0),\ (0,4),\ (0,k). $$Solution:
Using the formula for the area of a triangle,
$$ \text{Area} = \frac12 \left| \begin{vmatrix} -2&0&1\\ 0&4&1\\ 0&k&1 \end{vmatrix} \right|. $$Since the area is $4$ square units,
$$ \frac12 \left| \begin{vmatrix} -2&0&1\\ 0&4&1\\ 0&k&1 \end{vmatrix} \right| =4. $$Expanding the determinant along the first row,
$$ \begin{aligned} \begin{vmatrix} -2&0&1\\ 0&4&1\\ 0&k&1 \end{vmatrix} &= -2 \begin{vmatrix} 4&1\\ k&1 \end{vmatrix} +0 + 1 \begin{vmatrix} 0&4\\ 0&k \end{vmatrix} \\[2mm] &= -2(4-k)+0 \\[2mm] &= 2k-8. \end{aligned} $$Therefore,
$$ \frac12|2k-8|=4. $$ $$ |2k-8|=8. $$Hence,
Case 1:
$$ 2k-8=8 $$ $$ 2k=16 $$ $$ k=8. $$Case 2:
$$ 2k-8=-8 $$ $$ 2k=0 $$ $$ k=0. $$Hence, the required values of $k$ are
$$ \boxed{k=0\ \text{or}\ k=8.} $$Question 4
Find the equation of the line joining the following points using determinants.
(i)
The given points are
$$ (1,2)\quad\text{and}\quad(3,6). $$Solution:
The equation of the line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is
$$ \begin{vmatrix} x&y&1\\ x_1&y_1&1\\ x_2&y_2&1 \end{vmatrix} =0. $$Substituting the given points, we get
$$ \begin{vmatrix} x&y&1\\ 1&2&1\\ 3&6&1 \end{vmatrix} =0. $$Expanding the determinant along the first row,
$$ \begin{aligned} &x \begin{vmatrix} 2&1\\ 6&1 \end{vmatrix} -y \begin{vmatrix} 1&1\\ 3&1 \end{vmatrix} + \begin{vmatrix} 1&2\\ 3&6 \end{vmatrix} =0 \\[2mm] &x(2-6)-y(1-3)+(6-6)=0 \\[2mm] &-4x+2y=0. \end{aligned} $$Dividing both sides by $2$,
$$ -2x+y=0. $$Therefore, the required equation of the line is
$$ \boxed{y=2x.} $$(ii)
The given points are
$$ (3,1)\quad\text{and}\quad(9,3). $$Solution:
Using the determinant form of the equation of a line,
$$ \begin{vmatrix} x&y&1\\ 3&1&1\\ 9&3&1 \end{vmatrix} =0. $$Expanding the determinant along the first row,
$$ \begin{aligned} &x \begin{vmatrix} 1&1\\ 3&1 \end{vmatrix} -y \begin{vmatrix} 3&1\\ 9&1 \end{vmatrix} + \begin{vmatrix} 3&1\\ 9&3 \end{vmatrix} =0 \\[2mm] &x(1-3)-y(3-9)+(9-9)=0 \\[2mm] &-2x+6y=0. \end{aligned} $$Dividing both sides by $2$,
$$ -x+3y=0. $$Therefore, the required equation of the line is
$$ \boxed{x-3y=0.} $$Question 5
If the area of the triangle is $35$ square units with vertices $(2,-6)$, $(5,4)$ and $(k,4)$, then find the value of $k$.
(A) $12$
(B) $-2$
(C) $-12,\,-2$
(D) $12,\,-2$
Solution:
Using the formula for the area of a triangle,
$$ \text{Area} = \frac12 \left| \begin{vmatrix} 2&-6&1\\ 5&4&1\\ k&4&1 \end{vmatrix} \right|. $$Since the area of the triangle is $35$ square units,
$$ \frac12 \left| \begin{vmatrix} 2&-6&1\\ 5&4&1\\ k&4&1 \end{vmatrix} \right| =35. $$Expanding the determinant along the first row,
$$ \begin{aligned} \begin{vmatrix} 2&-6&1\\ 5&4&1\\ k&4&1 \end{vmatrix} &= 2 \begin{vmatrix} 4&1\\ 4&1 \end{vmatrix} -(-6) \begin{vmatrix} 5&1\\ k&1 \end{vmatrix} + \begin{vmatrix} 5&4\\ k&4 \end{vmatrix} \\[2mm] &= 2(4-4) +6(5-k) +(20-4k) \\[2mm] &= 0+30-6k+20-4k \\[2mm] &= 50-10k. \end{aligned} $$Therefore,
$$ \frac12|50-10k|=35. $$ $$ |50-10k|=70. $$Case 1:
$$ 50-10k=70 $$ $$ -10k=20 $$ $$ k=-2. $$Case 2:
$$ 50-10k=-70 $$ $$ -10k=-120 $$ $$ k=12. $$Hence, the required values of $k$ are
$$ \boxed{k=12\ \text{or}\ -2.} $$Correct Answer: (D) $12,\,-2$.