Exercise – 4.3
1. Find the area of the triangle with vertices at the points given in each of the following:
Solution:
-
$(1, 0), (6, 0), (4, 3)$
Area of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}1&0&1\\6&0&1\\4&3&1\end{array}} \right|$
$= \frac{1}{2}\left[ {1\left| {\begin{array}{*{20}{r}}0&1\\3&1\end{array}} \right| – 0 + 1\left| {\begin{array}{*{20}{r}}6&0\\4&3\end{array}} \right|} \right]$
$= \frac{1}{2}[1(0 – 3) + 1(18 – 0)] = \frac{15}{2} = 7.5$ sq. units. -
$(2, 7), (1, 1), (10, 8)$
Area of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}2&7&1\\1&1&1\\10&8&1\end{array}} \right|$
$= \frac{1}{2}\left[ {2\left| {\begin{array}{*{20}{r}}1&1\\8&1\end{array}} \right| – 7\left| {\begin{array}{*{20}{r}}1&1\\10&1\end{array}} \right| + 1\left| {\begin{array}{*{20}{r}}1&1\\10&8\end{array}} \right|} \right]$
$= \frac{1}{2}[2(1 – 8) – 7(1 – 10) + 1(8 – 10)] = \frac{47}{2} = 23.5$ sq. units. -
$(-2, -3), (3, 2), (-1, -8)$
Area of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}-2&-3&1\\3&2&1\\-1&-8&1\end{array}} \right|$
$= \frac{1}{2}\left[ {-2\left| {\begin{array}{*{20}{r}}2&1\\-8&1\end{array}} \right| + 3\left| {\begin{array}{*{20}{r}}3&1\\-1&1\end{array}} \right| + 1\left| {\begin{array}{*{20}{r}}3&2\\-1&-8\end{array}} \right|} \right]$
$= \frac{1}{2}[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)] = \frac{-30}{2} = -15$
Area $= 15$ square units. (Absolute value)
2. Show that points $A(a, b + c), B(b, c + a), C(c, a + b)$ are collinear.
Solution:
Consider $\left| {\begin{array}{*{20}{r}}a&b+c&1\\b&c+a&1\\c&a+b&1\end{array}} \right|$
Applying $C_1 \rightarrow C_1 + C_2$, we get
$= \left| {\begin{array}{*{20}{r}}a+b+c&b+c&1\\a+b+c&c+a&1\\a+b+c&a+b&1\end{array}} \right|$
Taking $(a + b + c)$ common from $C_1$, we get
$= (a + b + c)\left| {\begin{array}{*{20}{r}}1&b+c&1\\1&c+a&1\\1&a+b&1\end{array}} \right|$
$= (a + b + c) \times 0 = 0$
Hence, the points are collinear.
3. Find values of k if the area of the triangle is 4 sq. units and vertices are:
Solution:
-
$(k, 0), (4, 0), (0, 2)$
Area of triangle $= 4$ sq. units
Area $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}k&0&1\\4&0&1\\0&2&1\end{array}} \right|$
$= \frac{1}{2}[k(0 – 2) – 0 + 1(8 – 0)] = -k + 4$
Now, $-k + 4 = \pm 4 \Rightarrow -k + 4 = 4$ or $-k + 4 = -4$
$\Rightarrow k = 0$ or $k = 8$
$\Rightarrow k = 0, 8$ -
$(-2, 0), (0, 4), (0, k)$
Area of triangle $= 4$ sq. units
Area $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}-2&0&1\\0&4&1\\0&k&1\end{array}} \right|$
$= \frac{1}{2}[-2(4 – k) – 0 + 1(0)] = -4 + k$
Now, $-4 + k = \pm 4 \Rightarrow -4 + k = 4$ or $-4 + k = -4$
$\Rightarrow k = 0$ or $k = 8$
$\Rightarrow k = 0, 8$
4. Find the equation of the line joining the points using determinants:
Solution:
-
$(1, 2)$ and $(3, 6)$
Equation of the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is $\left| {\begin{array}{*{20}{r}}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}} \right| = 0$
$\Rightarrow \left| {\begin{array}{*{20}{r}}x&y&1\\1&2&1\\3&6&1\end{array}} \right| = 0$
$\Rightarrow x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0$
$\Rightarrow -4x + 2y = 0 \Rightarrow 2x – y = 0$
Hence, $y = 2x$ is the required line. -
$(3, 1)$ and $(9, 3)$
Equation of the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is $\left| {\begin{array}{*{20}{r}}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}} \right| = 0$
$\Rightarrow \left| {\begin{array}{*{20}{r}}x&y&1\\3&1&1\\9&3&1\end{array}} \right| = 0$
$\Rightarrow x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0 \Rightarrow -2x + 6y = 0$
$\Rightarrow -x + 3y = 0 \Rightarrow x – 3y = 0$
Hence, $x – 3y = 0$ is the required line.
5. If the area of a triangle is 35 sq. units with vertices $(2, -6), (5, 4)$ and $(k, 4)$, then k is:
Solution:
(D) 12, -2
Area of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}2&-6&1\\5&4&1\\k&4&1\end{array}} \right| = \pm 35$
$\Rightarrow \frac{1}{2}[2(4 – 4) + 6(5 – k) + 1(20 – 4k)] = \pm 35$
$\Rightarrow \frac{1}{2}[30 – 6k + 20 – 4k] = \pm 35$
$\Rightarrow \frac{1}{2}[50 – 10k] = \pm 35 \Rightarrow 25 – 5k = \pm 35$
$\Rightarrow 25 – 5k = 35$ or $25 – 5k = -35$
$\Rightarrow 5k = 10$ or $5k = 60$
$\Rightarrow k = -2$ or $k = 12$