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NCERT Solutions · Class 12 Mathematics Determinants

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EXERCISE 4.2 1–5 (part 2 of 8)

  1. Exercise 1

    Find area of the triangle with vertices at the point given in each of the following:
    (i)
    (1,0)\displaystyle (1, 0), (6,0)\displaystyle (6, 0), (4,3)\displaystyle (4,3)
    (ii)
    (2,7)\displaystyle (2, 7), (1,1)\displaystyle (1, 1), (10,8)\displaystyle (10,8)
    (iii)
    (−2,−3)\displaystyle (-2, -3), (3,2)\displaystyle (3, 2), (−1,−8)\displaystyle (-1, -8)

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{15}{2}\), (ii) \(\displaystyle \frac{47}{2}\), (iii) $\displaystyle 15$
    The area of the triangle with vertices \(\displaystyle (x_1,y_1),(x_2,y_2),(x_3,y_3)\) is \(\displaystyle \tfrac12\,|\Delta|\), where
    \[\Delta=\left|\begin{array}{rrr} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right|.\]
    The modulus sign is not decoration: the determinant itself can come out negative (it does in (iii)), while an area cannot.
    (i)
    Vertices \(\displaystyle (1,0),(6,0),(4,3)\). Expanding along \(\displaystyle R_1\),
    \[\Delta=\left|\begin{array}{rrr} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{array}\right| = 1(0\cdot 1-1\cdot 3)-0(6\cdot 1-1\cdot 4)+1(6\cdot 3-0\cdot 4) = -3+0+18 = 15.\]
    Area \(\displaystyle =\tfrac12|15| = \dfrac{15}{2}\) sq units.
    (ii)
    Vertices \(\displaystyle (2,7),(1,1),(10,8)\).
    \[\Delta=\left|\begin{array}{rrr} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{array}\right| = 2(1-8)-7(1-10)+1(8-10) = -14+63-2 = 47.\]
    Area \(\displaystyle =\tfrac12|47| = \dfrac{47}{2}\) sq units.
    (iii)
    Vertices \(\displaystyle (-2,-3),(3,2),(-1,-8)\). Watch the signs of the negative coordinates:
    \[\Delta=\left|\begin{array}{rrr} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{array}\right| = -2\big(2-(-8)\big)-(-3)\big(3-(-1)\big)+1\big(3(-8)-2(-1)\big),\]
    \[\Delta = -2(10)+3(4)+(-24+2) = -20+12-22 = -30.\]
    Area \(\displaystyle =\tfrac12|-30| = 15\) sq units.
    Answer: (i) \(\displaystyle \dfrac{15}{2}\) sq units, (ii) \(\displaystyle \dfrac{47}{2}\) sq units, (iii) \(\displaystyle 15\) sq units.

  2. Exercise 2

    Show that points A(a,b+c),B(b,c+a),C(c,a+b) are collinear. \mathrm{A}(a, b+c), \mathrm{B}(b, c+a), \mathrm{C}(c, a+b) \text { are collinear. }

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    Three points are collinear precisely when the triangle they determine has zero area, i.e. when the area determinant vanishes. So it is enough to show \[\Delta=\left|\begin{array}{rrr} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array}\right| = 0.\] Use the property that a determinant is unchanged if a multiple of one column is added to another; apply \(\displaystyle C_1 \to C_1 + C_2\): \[\Delta = \left|\begin{array}{rrr} a+b+c & b+c & 1 \\ a+b+c & c+a & 1 \\ a+b+c & a+b & 1 \end{array}\right|.\] Every entry of \(\displaystyle C_1\) now carries the common factor \(\displaystyle a+b+c\), which may be taken outside the determinant: \[\Delta = (a+b+c)\left|\begin{array}{rrr} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1 \end{array}\right|.\] In the remaining determinant \(\displaystyle C_1\) and \(\displaystyle C_3\) are identical, and a determinant with two identical columns is zero. Hence \[\Delta = (a+b+c)\cdot 0 = 0.\] Since the area \(\displaystyle \tfrac12|\Delta| = 0\), the points \(\displaystyle \mathrm{A}(a,b+c)\), \(\displaystyle \mathrm{B}(b,c+a)\), \(\displaystyle \mathrm{C}(c,a+b)\) are collinear. (Note the conclusion holds for all \(\displaystyle a,b,c\) — no need to assume \(\displaystyle a+b+c \neq 0\), because the second factor is zero on its own.)

  3. Exercise 3

    Find values of k\displaystyle k if area of triangle is 4\displaystyle 4 sq. units and vertices are
    (i)
    ( k,0),(4,0),(0,2)\displaystyle k, 0),(4,0),(0,2)
    (ii)
    (−2,0)\displaystyle (-2, 0), (0,4)\displaystyle (0, 4), (0\displaystyle 0, k)

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    NCERT’s answer
    (i)
    $\displaystyle 0$, $\displaystyle 8$, (ii) $\displaystyle 0$, $\displaystyle 8$
    With \(\displaystyle \Delta\) the area determinant, area \(\displaystyle =\tfrac12|\Delta| = 4\) gives \(\displaystyle |\Delta| = 8\). This is the step usually lost: the modulus splits into two equations, \(\displaystyle \Delta = 8\) and \(\displaystyle \Delta = -8\), and both must be solved — that is why each part has two answers.
    (i)
    Vertices \(\displaystyle (k,0),(4,0),(0,2)\). Expanding along \(\displaystyle R_1\),
    \[\Delta=\left|\begin{array}{rrr} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array}\right| = k(0\cdot 1-1\cdot 2)-0(4-0)+1(4\cdot 2-0) = -2k+8.\]
    So \(\displaystyle |8-2k| = 8\), giving
    \[8-2k = 8 \quad\Rightarrow\quad k = 0, \qquad\text{or}\qquad 8-2k = -8 \quad\Rightarrow\quad k = 8.\]
    (ii)
    Vertices \(\displaystyle (-2,0),(0,4),(0,k)\).
    \[\Delta=\left|\begin{array}{rrr} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right| = -2(4-k)-0(0-0)+1(0\cdot k-4\cdot 0) = 2k-8.\]
    So \(\displaystyle |2k-8| = 8\), giving
    \[2k-8 = 8 \quad\Rightarrow\quad k = 8, \qquad\text{or}\qquad 2k-8 = -8 \quad\Rightarrow\quad k = 0.\]
    Answer: \(\displaystyle k = 0\) or \(\displaystyle k = 8\) in both (i) and (ii).

  4. Exercise 4

    (i)
    Find equation of line joining (1,2)\displaystyle (1,2) and (3,6)\displaystyle (3,6) using determinants.
    (ii)
    Find equation of line joining (3,1)\displaystyle (3,1) and (9,3)\displaystyle (9,3) using determinants.

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    NCERT’s answer
    (i)
    \(\displaystyle y=2 x\), (ii) \(\displaystyle x-3 y=0\)
    A point \(\displaystyle (x,y)\) lies on the line through \(\displaystyle (x_1,y_1)\) and \(\displaystyle (x_2,y_2)\) exactly when the three points are collinear, i.e. when the area determinant is zero. So the equation of the line is
    \[\left|\begin{array}{rrr} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{array}\right| = 0.\]
    (i)
    Through \(\displaystyle (1,2)\) and \(\displaystyle (3,6)\):
    \[\left|\begin{array}{rrr} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{array}\right| = 0.\]
    Expanding along \(\displaystyle R_1\),
    \[x(2\cdot 1-1\cdot 6)-y(1\cdot 1-1\cdot 3)+1(1\cdot 6-2\cdot 3) = 0,\]
    \[-4x+2y+0 = 0 \quad\Rightarrow\quad 2y = 4x.\]
    Equation of the line: \(\displaystyle y = 2x\).
    (ii)
    Through \(\displaystyle (3,1)\) and \(\displaystyle (9,3)\):
    \[\left|\begin{array}{rrr} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{array}\right| = 0.\]
    Expanding along \(\displaystyle R_1\),
    \[x(1\cdot 1-1\cdot 3)-y(3\cdot 1-1\cdot 9)+1(3\cdot 3-1\cdot 9) = 0,\]
    \[-2x+6y+0 = 0 \quad\Rightarrow\quad x = 3y.\]
    Equation of the line: \(\displaystyle x - 3y = 0\), i.e. \(\displaystyle x = 3y\).

  5. Exercise 5

    If area of triangle is 35\displaystyle 35 sq units with vertices (2,−6),(5,4)\displaystyle (2,-6),(5,4) and (k,4)\displaystyle (k, 4). Then k\displaystyle k is (A) 12\displaystyle 12 (B) -2\displaystyle 2 (C) -12\displaystyle 12, -2\displaystyle 2 (D) 12\displaystyle 12,-2\displaystyle 2

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    NCERT’s answer
    (D)
    Form the area determinant for \(\displaystyle (2,-6),(5,4),(k,4)\) and expand along \(\displaystyle R_1\): \[\Delta=\left|\begin{array}{rrr} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right| = 2(4-4)-(-6)(5-k)+1(5\cdot 4-4k) = 0+6(5-k)+(20-4k),\] \[\Delta = 30-6k+20-4k = 50-10k.\] Area \(\displaystyle =\tfrac12|\Delta| = 35\), so \(\displaystyle |50-10k| = 70\). Keeping both branches of the modulus: \[50-10k = 70 \quad\Rightarrow\quad k = -2, \qquad\text{or}\qquad 50-10k = -70 \quad\Rightarrow\quad k = 12.\] Both values are admissible (neither makes the three points collinear), so \(\displaystyle k = 12\) or \(\displaystyle k = -2\).Answer: option (D) \(\displaystyle 12,\,-2\).