12th NCERT Matrices Exercise 4.3 Questions 5
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Question (1)

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

Solution

(i) (1, 0), (6, 0), (4, 3)
$D = \left| {\begin{array}{*{20}{c}}1&0&1\\6&0&1\\4&3&1\end{array}} \right|$ $D = 1\left( {0 - 3} \right) - 0 + 1\left( {18 - 0} \right)$ $D = - 3 + 18 = 15$ $Area\quad \Delta = \frac{1}{2}\left| D \right|$ $Area = \frac{1}{2}\left| {15} \right| = \frac{{15}}{2}$
(ii) (2, 7), (1, 1), (10, 8)
$D = \left| {\begin{array}{*{20}{c}}2&7&1\\1&1&1\\{10}&8&1\end{array}} \right|$ $D = 2\left( {1 - 8} \right) - 7\left( {1 - 10} \right) + 1\left( {8 - 10} \right)$ $D = - 14 + 63 - 2 = 47$ $Area\;\Delta = \frac{1}{2}\left| D \right|$
(iii) (-2, -3) (3, 2) (-1, 8)
$D = \left| {\begin{array}{*{20}{c}}{ - 2}&{ - 3}&1\\3&2&1\\{ - 1}&{ - 8}&1\end{array}} \right|$ $D = - 2\left( {2 + 8} \right) + 3\left( {3 + 1} \right) + 1\left( { - 24 + 2} \right)$ $D = - 20 + 12 - 22 = - 30$ $Area\;\Delta = \frac{1}{2}\left| D \right|$ $Area\;\Delta = \frac{1}{2}\left| { - 30} \right| = 15$

Question (2)

Show that points
A(a, b+c), B(b, c+a), C(c, a+b) are collinear

Solution

$D = \left| {\begin{array}{*{20}{c}}a&{b + c}&1\\b&{c + a}&1\\c&{a + b}&1\end{array}} \right|$ ${C_2} \to {C_2} + {C_1}$ $\left| {\begin{array}{*{20}{c}}a&{a + b + c}&1\\b&{b + c + a}&1\\c&{c + a + b}&1\end{array}} \right|$ ${C_2}\left( {\frac{1}{{a + b + c}}} \right)$ $= \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}a&1&1\\b&1&1\\c&1&1\end{array}} \right|$ $\text{As} \quad {C_2} = {C_1}$ $D = \left( {a + b + c} \right)\left( 0 \right) = 0$ Since D = 0, points are collinear

Question (3)

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

Solution

Find k if Area = 4
$Area\;\Delta = \frac{1}{2}\left| D \right|$ $4 = \frac{1}{2}\left| D \right|$ $\left| D \right| = 8$ $D = \left| {\begin{array}{*{20}{c}}k&0&1\\4&0&1\\0&2&1\end{array}} \right|$ $D = k\left( {0 - 2} \right) - 0 + 1\left( {8 - 0} \right)$ $D = 2k - 8$ As |D| = 8
$8 = \left| { - 2k + 8} \right|$ $- 2k + 8 = \pm 8$ $If\;2k - 8 = + 8$ $k = 8$ $If\;2k - 8 = - 8$ $k = 0$ Therefore k=8 or 0

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) A(1, 2) B (3, 6)
Equation of line $\overleftrightarrow {xy}$ is given by
$\left| {\begin{array}{*{20}{c}}x&y&1\\1&2&1\\3&6&1\end{array}} \right| = 0$ $x\left( {2 - 6} \right) - y\left( {1 - 3} \right) + 1\left( {6 - 6} \right) = 0$ $- 4x + 2y = 0$ $2x - y = 0$
(ii) A(3, 1), B(9, 3)
Equation of $\overleftrightarrow {AB}$ is given by
$\left| {\begin{array}{*{20}{c}}x&y&1\\3&1&1\\9&3&1\end{array}} \right| = 0$ $x\left( {1 - 3} \right) - y\left( {3 - 9} \right) + 1\left( {9 - 9} \right) = 0$ $- 2x + 6y = 0$ $x - 3y = 0$

Question (5)

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

Solution

Area = 35, A(2, -6), B(5, 4) C(k, 4)
$Area\;\Delta = \frac{1}{2}\left| D \right|$ $35 = \frac{1}{2}\left| D \right|$ $\left| D \right| = 70$ $70 = \left| {\begin{array}{*{20}{c}}2&{ - 6}&1\\5&4&1\\k&4&1\end{array}} \right|$ $= 2\left( {4 - 4} \right) + 6\left( {5 - k} \right) + 1\left( {20 - 4k} \right)$ $= 30 - 6k + 20 - 4k$ $= 50 - 10k$ $\left| {50 - 10k} \right| = 70$ $50 - 10k = \pm 70$ $k = - 2$ $50 - 10k = - 70$ $k = 12$ There4; k=12, -2 so option 'D' is correct answer