Class 12 ncert solution math exercise 4.3 determinants

   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

It is given that the area of the triangle is 35 square units

Hence, \Delta=\pm 35.

Therefore,

\Rightarrow 25-5 k=\pm 35
\Rightarrow 5(5-k)=\pm 35
\Rightarrow 5-k=\pm 7

When, 5-k=-7

Then, k=12

When, 5-k=7

Then, k=-2

Hence, k=12,-2

Thus, the correct option is D.

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