EXERCISE 4.3 (Determinants)

Question 1: Find area of the triangle with vertices at the point given in each of the following:(Class 12 ncert solution math exercise 4.3 determinants)

(i) $(1,0),(6,0),(4,3)$

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

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

Solution: (i) The area of the triangle with vertices $(1,0),(6,0),(4,3)$ is given by the relation,

$\Delta =\frac{1}{2}\left|\begin{array}{lll} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{array}\right|$

$=\frac{1}{2}[1(0-3)-0(6-4)+1(18-0)]$
$=\frac{1}{2}[-3+18]$
$=\frac{1}{2}[15]$

$=\frac{15}{2}$

Hence, area of the triangle is $\frac{15}{2}$ square units.

(ii) The area of the triangle with vertices $(2,7),(1,1),(10,8)$ is given by the relation,

$\Delta =\frac{1}{2}\left|\begin{array}{ccc} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{array}\right|$
$=\frac{1}{2}[2(1-8)-7(1-10)+1(8-10)]$
$=\frac{1}{2}[2(-7)-7(-9)+1(-2)]$
$=\frac{1}{2}[-14+63-2]$
$=\frac{1}{2}[47]$
$=\frac{47}{2}$

Hence, area of the triangle is $\frac{47}{2}$ square units.

(iii) The area of the triangle with vertices $(-2,-3),(3,2),(-1,-8)$ is given by the relation,

$\Delta =\frac{1}{2}\left|\begin{array}{ccc} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{array}\right|$

$=\frac{1}{2}[-2(2+8)+3(3+1)+1(-24+2)]$
$=\frac{1}{2}[-2(10)+3(4)+1(-22)]$
$=\frac{1}{2}[-20+12-22]$
$=-\frac{1}{2}[30]$
$=-15$

Hence, area of the triangle is 15 square units.

Question 2: Show that the points $A(a, b+c), B(b, c+a), C(c, a+b)$ are collinear.

Solution: The area of the triangle with vertices $A(a, b+c), B(b, c+a), C(c, a+b)$ is given by the absolute value of the relation:

$\Delta =\frac{1}{2}\left|\begin{array}{lll} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc} a & b+c & 1 \\ b-a & a-b & 0 \\ c-a & a-c & 0 \end{array}\right|$

$=\frac{1}{2}(a-b)(c-a)\left|\begin{array}{ccc} a & b+c & 1 \\ -1 & 1 & 0 \\ 1 & -1 & 0 \end{array}\right|$

$=\frac{1}{2}(a-b)(c-a)\left|\begin{array}{ccc} a & b+c & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right|$
$=0$

Thus, the area of the triangle formed by points is zero.

Hence, the points are collinear.

Question 3: Find values of $k$ if area of triangle is 4 square units and vertices are:

(i) $(k, 0),(4,0),(0,2)$

(ii) $(-2,0),(0,4),(0, k)$

Solution: We know that the area of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is the absolute value of the determinant $(\Delta)$ , where

$\Delta=\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|$

It is given that the area of triangle is 4 square units.

Hence, $\Delta=\pm 4$

(i) The area of the triangle with vertices $(k, 0),(4,0),(0,2)$ is given by the relation,

$\Delta =\frac{1}{2}\left|\begin{array}{lll} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array}\right|$

$=\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)]$
$=\frac{1}{2}[-2 k+8]$
$=-k+4$

Therefore, $-k+4=\pm 4$

When $-k+4=-4$

Then $k=8$

When $-k+4=4$

Then $k=0$

Hence, $k=0,8$

(ii) The area of the triangle with vertices $(-2,0),(0,4),(0, k)$ is given by the relation,

$\Delta =\frac{1}{2}\left|\begin{array}{ccc} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right|$

$=\frac{1}{2}[-2(4-k)]$
$=k-4$

Therefore, $-k+4=\pm 4$

When $k-4=4$

Then $k=8$

When $k-4=-4$

Then $k=0$

Hence, $k=0,8$

Question 4: (i) Find equation of line joining $(1,2)$ and $(3,6)$ using determinants.

(ii) Find equation of line joining $(3,1)$ and $(9,3)$ using determinants.

Solution: (i) Let $P(x, y)$ be any point on the line joining points $A(1,2)$ and $B(3,6)$ .

Then, the points $A, B$ and $P$ are collinear.

Hence, the area of triangle $A B P$ will be zero.

Therefore,

$\Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{array}\right|=0$
$\Rightarrow \frac{1}{2}[1(6-y)-2(3-x)+1(3 y-6 x)]=0$
$\Rightarrow 6-y-6+2 x+3 y-6 x=0$
$\Rightarrow y=2 x$

Thus, the equation of the line joining the given points is $y=2 x$ .

(ii) Let $P(x, y)$ be any point on the line joining points $A(3,1)$ and $B(9,3)$ .

Then, the points $A, B$ and $P$ are collinear.

Hence, the area of triangle $A B P$ will be zero.

Therefore,

$\Rightarrow \frac{1}{2}\left|\begin{array}{lll} 3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1 \end{array}\right|=0$
$\Rightarrow \frac{1}{2}[3(3-y)-1(9-x)+1(9 y-3 x)]=0$
$\Rightarrow 9-3 y-9+x+9 y-3 x=0$
$\Rightarrow 6 y-2 x=0$
$\Rightarrow x-3 y=0$

Thus, the equation of the line joining the given points is $x-3 y=0$ .

Question 5: If area of the triangle is 35 square units with vertices $(2,-6),(5,4),(k, 4)$ . Then $k$ is
(A) 12
(B) $-2$
(C) $-12,-2$
(D) $12,-2$

Solution: The area of the triangle with vertices $(2,-6),(5,4),(k, 4)$ is given by the relation,

$\Delta =\frac{1}{2}\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|$
$=\frac{1}{2}[2(4-4)+6(5-k)+1(20-4 k)]$
$=\frac{1}{2}[30-6 k+20-4 k]$
$=\frac{1}{2}[50-10 k]$
$=25-5 k$