NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.3

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Chapter 4 Ex.4.3 Question 1

Find area of the triangle with vertices at the point given in each of the following:

(i) \(\left( {1,0} \right),\left( {6,0} \right),\left( {4,3} \right)\)

(ii) \(\left( {2,7} \right),\left( {1,1} \right),\left( {10,8} \right)\)

(iii) \(\left( { - 2, - 3} \right),\left( {3,2} \right),\left( { - 1, - 8} \right)\)

Solution

(i) The area of the triangle with vertices \(\left( {1,0} \right),\left( {6,0} \right),\left( {4,3} \right)\) is given by the relation,

\[\begin{align}\Delta&= \frac{1}{2}\left| {\begin{array}{*{20}{c}}1&0&1\\6&0&1\\4&3&1\end{array}} \right|\\&= \frac{1}{2}\left[ {1\left( {0 - 3} \right) - 0\left( {6 - 4} \right) + 1\left( {18 - 0} \right)} \right]\\&= \frac{1}{2}\left[ { - 3 + 18} \right]\\&= \frac{1}{2}\left[ {15} \right]\\&= \frac{{15}}{2}\end{align}\]

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

(ii) The area of the triangle with vertices \(\left( {2,7} \right),\left( {1,1} \right),\left( {10,8} \right)\) is given by the relation,

\[\begin{align}\Delta&= \frac{1}{2}\left| {\begin{array}{*{20}{c}}2&7&1\\1&1&1\\{10}&8&1\end{array}} \right|\\&= \frac{1}{2}\left[ {2\left( {1 - 8} \right) - 7\left( {1 - 10} \right) + 1\left( {8 - 10} \right)} \right]\\&= \frac{1}{2}\left[ {2\left( { - 7} \right) - 7\left( { - 9} \right) + 1\left( { - 2} \right)} \right]\\&= \frac{1}{2}\left[ { - 14 + 63 - 2} \right]\\&= \frac{1}{2}\left[ {47} \right]\\&= \frac{{47}}{2}\end{align}\]

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

(iii) The area of the triangle with vertices \(\left( { - 2, - 3} \right),\left( {3,2} \right),\left( { - 1, - 8} \right)\) is given by the relation,

\[\begin{align}\Delta&= \frac{1}{2}\left| {\begin{array}{*{20}{c}}{ - 2}&{ - 3}&1\\3&2&1\\{ - 1}&{ - 8}&1\end{array}} \right|\\&= \frac{1}{2}\left[ { - 2\left( {2 + 8} \right) + 3\left( {3 + 1} \right) + 1\left( { - 24 + 2} \right)} \right]\\&= \frac{1}{2}\left[ { - 2\left( {10} \right) + 3\left( 4 \right) + 1\left( { - 22} \right)} \right]\\&= \frac{1}{2}\left[ { - 20 + 12 - 22} \right]\\&=- \frac{1}{2}\left[ {30} \right]\\&=- 15\end{align}\]

Hence, area of the triangle is \(15\) square units.

Chapter 4 Ex.4.3 Question 2

Show that the points \(A\left( {a,b + c} \right),B\left( {b,c + a} \right),C\left( {c,a + b} \right)\) are collinear.

Solution

The area of the triangle with vertices \(A\left( {a,b + c} \right),B\left( {b,c + a} \right),C\left( {c,a + b} \right)\) is given by the absolute value of the relation:

\[\begin{align}\Delta&= \frac{1}{2}\left| {\begin{array}{*{20}{c}}a&{b + c}&1\\b&{c + a}&1\\c&{a + b}&1\end{array}} \right|\\&= \frac{1}{2}\left| {\begin{array}{*{20}{c}}a&{b + c}&1\\{b - a}&{a - b}&0\\{c - a}&{a - c}&0\end{array}} \right|\qquad \left[ {{R_2} \to {R_2} - {R_1}{\text{ and }}{R_3} \to {R_3} - {R_1}} \right]\\&= \frac{1}{2}\left( {a - b} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}a&{b + c}&1\\{ - 1}&1&0\\1&{ - 1}&0\end{array}} \right|\\&= \frac{1}{2}\left( {a - b} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}a&{b + c}&1\\{ - 1}&1&0\\0&0&0\end{array}} \right| \qquad \left[ {{R_3} \to {R_3} + {R_2}} \right]\\&= 0\end{align}\]

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

Hence, the points are collinear.

Chapter 4 Ex.4.3 Question 3

Find values of \(k\) if area of triangle is 4 square units and vertices are:

(i) \(\left( {k,0} \right),\left( {4,0} \right),\left( {0,2} \right)\)

(ii) \(\left( { - 2,0} \right),\left( {0,4} \right),\left( {0,k} \right)\)

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 \(~\left( \Delta\right)\), where

\[\Delta= \frac{1}{2}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|\]

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

Hence, \(\Delta=\pm 4\)

(i) The area of the triangle with vertices \(\left( {k,0} \right),\left( {4,0} \right),\left( {0,2} \right)\) is given by the relation,

\[\begin{align}\Delta&= \frac{1}{2}\left| {\begin{array}{*{20}{c}}k&0&1\\4&0&1\\0&2&1\end{array}} \right|\\&= \frac{1}{2}\left[ {k\left( {0 - 2} \right) - 0\left( {4 - 0} \right) + 1\left( {8 - 0} \right)} \right]\\&= \frac{1}{2}\left[ { - 2k + 8} \right]\\&=- k + 4\end{align}\]

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 \(\left( { - 2,0} \right),\left( {0,4} \right),\left( {0,k} \right)\) is given by the relation,

\[\begin{align}\Delta&= \frac{1}{2}\left| {\begin{array}{*{20}{c}}{ - 2}&0&1\\0&4&1\\0&k&1\end{array}} \right|\\&= \frac{1}{2}\left[ { - 2\left( {4 - k} \right)} \right]\\&= k - 4\end{align}\]

Therefore, \( - k + 4 =\pm 4\)

When \(k - 4 = 4\)

Then \(k = 8\)

When \(k - 4 =- 4\)

Then \(k = 0\)

Hence, \(k = 0,8\)

Chapter 4 Ex.4.3 Question 4

(i) Find equation of line joining \(\left( {1,2} \right)\) and \(\left( {3,6} \right)\) using determinants.

(ii) Find equation of line joining \(\left( {3,1} \right)\) and \(\left( {9,3} \right)\) using determinants.

Solution

(i) Let \(P\left( {x,y} \right)\) be any point on the line joining points \(A\left( {1,2} \right)\) and \(B\left( {3,6} \right)\).

Then, the points \(A,B\) and \(P\) are collinear.

Hence, the area of triangle \(ABP\) will be zero.

Therefore,

\[\begin{align}&\Rightarrow \;\frac{1}{2}\left| {\begin{array}{*{20}{c}}1&2&1\\3&6&1\\x&y&1\end{array}} \right| = 0\\&\Rightarrow\; \frac{1}{2}\left[ {1\left( {6 - y} \right) - 2\left( {3 - x} \right) + 1\left( {3y - 6x} \right)} \right] = 0\\&\Rightarrow\; 6 - y - 6 + 2x + 3y - 6x = 0\\&\Rightarrow\; 2y - 4x = 0\\&\Rightarrow\; y = 2x\end{align}\]

Thus, the equation of the line joining the given points is \(y = 2x\).

(ii) Let \(P\left( {x,y} \right)\) be any point on the line joining points \(A\left( {3,1} \right)\) and \(B\left( {9,3} \right)\).

Then, the points \(A,B\) and \(P\) are collinear.

Hence, the area of triangle \(ABP\) will be zero.

Therefore,

\[\begin{align}&\Rightarrow\; \frac{1}{2}\left| {\begin{array}{*{20}{c}}3&1&1\\9&3&1\\x&y&1\end{array}} \right| = 0\\&\Rightarrow \;\frac{1}{2}\left[ {3\left( {3 - y} \right) - 1\left( {9 - x} \right) + 1\left( {9y - 3x} \right)} \right] = 0\\&\Rightarrow\; 9 - 3y - 9 + x + 9y - 3x = 0\\&\Rightarrow\; 6y - 2x = 0\\&\Rightarrow\; x - 3y = 0\end{align}\]

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

Chapter 4 Ex.4.3 Question 5

If area of the triangle is \(35\) square units with vertices\(\left( {2, - 6} \right),\left( {5,4} \right),\left( {k,4} \right)\). Then \(k\) is

(A) \(12\)

(B) \( - 2\)

(C) \( - 12, - 2\)

(D) \(12, - 2\)

Solution

The area of the triangle with vertices \(\left( {2, - 6} \right),\left( {5,4} \right),\left( {k,4} \right)\) is given by the relation,

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

It is given that the area of the triangle is \(35\) square units

Hence, \(\Delta=\pm {\rm{35}}\).

Therefore,

\[\begin{align}&\Rightarrow\; 25 - 5k =\pm 35\\&\Rightarrow\; 5\left( {5 - k} \right) =\pm 35\\&\Rightarrow \;5 - k =\pm 7\end{align}\]

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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