OP Malhotra Class 12 Maths Solutions Chapter 5 Determinants Exercise 5 (C)

Question 1. Find the area of the triangle whose vertices are :
(i) (- 8, - 2), (- 4, - 6), (- 1, 5)
(ii) (- 3, 5), (3, - 6), (7, 2)
Answer:
(i) The area of a triangle with given vertices is found using the determinant formula. We use the coordinates (- 8, - 2), (- 4, - 6), and (- 1, 5).
\[ \text{Required Area} = \frac{1}{2} \left| \begin{array}{ccc} -8 & -2 & 1 \\ -4 & -6 & 1 \\ -1 & 5 & 1 \end{array} \right| \] Expanding along the first row (R1), we get:
\[ = \left| \frac{1}{2} \{-8(-6-5) - (-2)(-4 - (-1)) + 1(-4(5) - (-6)(-1))\} \right| \] \[ = \left| \frac{1}{2} \{-8(-11) + 2(-4+1) + 1(-20-6)\} \right| \] \[ = \left| \frac{1}{2} \{88 + 2(-3) + 1(-26)\} \right| \] \[ = \left| \frac{1}{2} \{88 - 6 - 26\} \right| \] \[ = \left| \frac{1}{2} \{56\} \right| \] \[ = 28 \text{ square units} \]
(ii) For the second set of coordinates (- 3, 5), (3, - 6), and (7, 2), we again use the determinant formula.
\[ \text{Required Area} = \frac{1}{2} \left| \begin{array}{ccc} -3 & 5 & 1 \\ 3 & -6 & 1 \\ 7 & 2 & 1 \end{array} \right| \] Expanding along the third row (R3), we get:
\[ = \left| \frac{1}{2} \{7(5(1) - (-6)(1)) - 2(-3(1) - 3(1)) + 1(-3(-6) - 5(3))\} \right| \] \[ = \left| \frac{1}{2} \{7(5+6) - 2(-3-3) + 1(18-15)\} \right| \] \[ = \left| \frac{1}{2} \{7(11) - 2(-6) + 1(3)\} \right| \] \[ = \left| \frac{1}{2} \{77 + 12 + 3\} \right| \] \[ = \left| \frac{1}{2} \{92\} \right| \] \[ = 46 \text{ square units} \]In simple words: To find the area of a triangle when you know its corner points, you use a special math tool called a determinant. You put the coordinates into a grid, do some calculations, and then divide the final number by two. The absolute value ensures the area is always positive.

🎯 Exam Tip: Remember to always take the absolute value of the result when calculating area, as area cannot be negative. Also, ensure you apply the \( \frac{1}{2} \) factor correctly.

Question 2. Using determinants, prove that the following points are collinear:
(i) (11, 7), (5, 5), (- 1, 3)
(ii) (0, 3), (4, 6), (- 8, - 3)
Answer: For points to be collinear (lie on the same straight line), the area of the triangle formed by them must be zero. We will calculate the area using determinants.
(i) For the points (11, 7), (5, 5), and (- 1, 3):
\[ \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} 11 & 7 & 1 \\ 5 & 5 & 1 \\ -1 & 3 & 1 \end{array} \right| \] Expanding along the third column (C3), we get:
\[ = \left| \frac{1}{2} \{1(5(3) - 5(-1)) - 1(11(3) - 7(-1)) + 1(11(5) - 7(5))\} \right| \] \[ = \left| \frac{1}{2} \{1(15+5) - 1(33+7) + 1(55-35)\} \right| \] \[ = \left| \frac{1}{2} \{20 - 40 + 20\} \right| \] \[ = \left| \frac{1}{2} \{0\} \right| \] \[ = 0 \text{ square units} \] Since the area is 0, the given points are collinear.
(ii) For the points (0, 3), (4, 6), and (- 8, - 3):
\[ \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} 0 & 3 & 1 \\ 4 & 6 & 1 \\ -8 & -3 & 1 \end{array} \right| \] Expanding along the third column (C3), we get:
\[ = \left| \frac{1}{2} \{1(4(-3) - 6(-8)) - 1(0(-3) - 3(-8)) + 1(0(6) - 3(4))\} \right| \] \[ = \left| \frac{1}{2} \{1(-12+48) - 1(0+24) + 1(0-12)\} \right| \] \[ = \left| \frac{1}{2} \{36 - 24 - 12\} \right| \] \[ = \left| \frac{1}{2} \{0\} \right| \] \[ = 0 \text{ square units} \] Since the area is 0, these given points are also collinear.In simple words: When three points lie on the same straight line, they cannot form a triangle. So, the "area" of a triangle made by these points will be zero. We use the determinant calculation to check if this area is zero, which proves they are on the same line.

🎯 Exam Tip: The key concept for collinearity using determinants is that the area of the triangle formed by the points must be zero. If you get a non-zero area, the points are not collinear.

Question 3. Show that the points (b, c + a), (c, a + b) and (a, b + c) are collinear.
Answer: To show that the given points are collinear, we must prove that the area of the triangle formed by them is zero. The points are \( (x_1, y_1) = (b, c+a) \), \( (x_2, y_2) = (c, a+b) \), and \( (x_3, y_3) = (a, b+c) \).
The area of the triangle is given by:
\[ \Delta = \frac{1}{2} \left| \begin{array}{ccc} b & c+a & 1 \\ c & a+b & 1 \\ a & b+c & 1 \end{array} \right| \] Expanding the determinant along the third column (C3), we get:
\[ \Delta = \frac{1}{2} \left| 1 \cdot (c(b+c) - a(a+b)) - 1 \cdot (b(b+c) - a(c+a)) + 1 \cdot (b(a+b) - c(c+a)) \right| \] Now, let's simplify the terms inside the absolute value:
The first term is: \( c(b+c) - a(a+b) = bc + c^2 - a^2 - ab \)
The second term is: \( b(b+c) - a(c+a) = b^2 + bc - ac - a^2 \)
The third term is: \( b(a+b) - c(c+a) = ab + b^2 - c^2 - ac \)
So, substituting these back:
\[ \Delta = \frac{1}{2} \left| (bc + c^2 - a^2 - ab) - (b^2 + bc - ac - a^2) + (ab + b^2 - c^2 - ac) \right| \] Now, let's combine and cancel terms within the brackets:
\[ \Delta = \frac{1}{2} \left| bc + c^2 - a^2 - ab - b^2 - bc + ac + a^2 + ab + b^2 - c^2 - ac \right| \] Many terms will cancel each other out:
\( (c^2 - c^2) + (a^2 - a^2) + (bc - bc) + (ab - ab) + (b^2 - b^2) + (ac - ac) = 0 \)

\[ \implies \Delta = \frac{1}{2} |0| \]
\[ \implies \Delta = 0 \] Since the area of the triangle is 0, the given points are collinear.In simple words: We calculated the area of the triangle formed by these three points using a special math grid. When we added and subtracted all the parts, the final answer was zero. This means the points don't form a triangle and must lie on the same straight line.

🎯 Exam Tip: When proving collinearity for algebraic coordinates, carefully expand each minor and watch for cancellation of terms. A common mistake is algebraic error during expansion or sign errors.

Question 4. Find x so that the points (3, - 2), (x, 2) and (8, 8) be on a line.
Answer: For three points to be on the same line (collinear), the area of the triangle formed by them must be zero. We will set up the determinant for the area and equate it to zero to find the value of x.
The points are (3, - 2), (x, 2), and (8, 8).
\[ \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} 3 & -2 & 1 \\ x & 2 & 1 \\ 8 & 8 & 1 \end{array} \right| \] Since the points are collinear, the area is 0:
\[ \frac{1}{2} \left| \begin{array}{ccc} 3 & -2 & 1 \\ x & 2 & 1 \\ 8 & 8 & 1 \end{array} \right| = 0 \] We can remove the \( \frac{1}{2} \) as it does not affect the determinant being zero. Expanding along the third column (C3):
\[ |1(x(8) - 2(8)) - 1(3(8) - (-2)(8)) + 1(3(2) - (-2)(x))| = 0 \] \[ |(8x - 16) - (24 + 16) + (6 + 2x)| = 0 \]
\[ \implies |8x - 16 - 40 + 6 + 2x| = 0 \]
\[ \implies |10x - 50| = 0 \] For the absolute value to be zero, the expression inside must be zero:

\[ \implies 10x - 50 = 0 \]
\[ \implies 10x = 50 \]
\[ \implies x = 5 \] The value of x is 5 for the points to be collinear.In simple words: When three points are on one line, they can't make a triangle, so the triangle's area is zero. We put the point coordinates into a special math grid and set its calculated "area" to zero. Then we solve the equation to find the missing 'x' value.

🎯 Exam Tip: When solving for an unknown variable like 'x' with collinear points, remember that the entire determinant (not just the determinant itself, but the entire area calculation including the 1/2 factor) must be set to zero. The absolute value bars become important only if the result before setting to zero is negative, but here for an equation, simply equate the expression to zero.

Question 5. If (x, y), (a, 0), (0, b) are collinear, then using determinants prove that \( \frac{x}{a}+\frac{y}{b} = 1 \).
Answer: If three points are collinear, the area of the triangle formed by them is zero. We will use this property for the given points (x, y), (a, 0), and (0, b).
The area of the triangle is:
\[ \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} x & y & 1 \\ a & 0 & 1 \\ 0 & b & 1 \end{array} \right| \] Since the points are collinear, the area must be 0:
\[ \frac{1}{2} \left| \begin{array}{ccc} x & y & 1 \\ a & 0 & 1 \\ 0 & b & 1 \end{array} \right| = 0 \] We can ignore the \( \frac{1}{2} \) as it won't change the equation being zero. Expanding the determinant along the third column (C3):
\[ |1(a(b) - 0(0)) - 1(x(b) - y(0)) + 1(x(0) - y(a))| = 0 \] \[ |(ab - 0) - (bx - 0) + (0 - ay)| = 0 \]
\[ \implies |ab - bx - ay| = 0 \] For the absolute value to be zero, the expression inside must be zero:

\[ \implies ab - bx - ay = 0 \] Rearranging the terms to match the required result:

\[ \implies bx + ay = ab \] Now, divide the entire equation by `ab` (assuming a and b are not zero, which they must be for the expression \( \frac{x}{a}+\frac{y}{b} \) to be defined):

\[ \implies \frac{bx}{ab} + \frac{ay}{ab} = \frac{ab}{ab} \]
\[ \implies \frac{x}{a} + \frac{y}{b} = 1 \] This is the required result.In simple words: We know that points on a straight line make a triangle with zero area. We set up the area formula with the given points and solved the equation. After some algebraic steps, we rearranged the result to show the specific relationship \( \frac{x}{a}+\frac{y}{b} = 1 \).

🎯 Exam Tip: Remember to clearly state the condition for collinearity (Area = 0) and show each step of the determinant expansion and algebraic manipulation. Dividing by `ab` at the end is a crucial step to reach the target identity.