GSEB Class 12 Maths Solutions Chapter 4 Determinants Exercise 4.3

Question 1. Find the area of the triangle with vertices at the points 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)
Answer:
(i) The area \( A \) of the triangle whose vertices are \( (1, 0), (6, 0), (4, 3) \) is calculated by the formula:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| = \frac { 1 }{ 2 }\left|\begin{array}{lll} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{array}\right| \)
When we expand using the elements of the second column, we get:
\( A = \frac { 1 }{ 2 }| 3 \times (1 – 6) | = \frac { 1 }{ 2 }| 3 \times (-5) | = \frac { 1 }{ 2 }|-15| = \frac { 15 }{ 2 } = 7 \frac { 1 }{ 2 } \) Sq. units.
(ii) The vertices of triangle ABC are \( A (2, 7), B (1, 1), C (10, 8) \).
The area \( A \) of triangle ABC is given by:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| = \frac { 1 }{ 2 }\left|\begin{array}{ccc} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{array}\right| \)
Expanding with the help of elements of the first column, we obtain:
\( A = \frac { 1 }{2}[2(1 – 8) – 1(7 – 8) + 10(7 – 1)] \)
\( = \frac { 1 }{ 2 } [2(-7) – 1(-1) + 10(6)] \)
\( = \frac { 1 }{ 2 } [-14 + 1 + 60] \)
\( = \frac { 47 }{ 2 } = 23\frac { 1 }{ 2 } \) Sq. units.
(iii) The vertices of triangle ABC are \( A (-2, -3), B (3, 2), C (-1, -8) \).
The area of triangle ABC is:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| = \frac { 1 }{ 2 }\left|\begin{array}{ccc} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{array}\right| \)
Expanding the determinant, we get:
\( = \frac { 1 }{ 2 }[-2(2 + 8) - (-3)(3 + 1) + 1(- 24 + (-1 \times 2))] \)
\( = \frac { 1 }{ 2 }[-2(10) + 3(4) + 1(- 24 - 2)] \)
\( = \frac { 1 }{ 2 }[-20 + 12 – 26] \)
\( = \frac { 1 }{ 2 }(-34) = -17 \).
Ignoring the negative sign, the area of triangle ABC is \( 17 \) sq. units.
In simple words: To find a triangle's area using its corner points, we use a special math tool called a determinant. You set up a grid with the x and y numbers of the points, plus a column of ones. Then you calculate a value from this grid. Half of that value gives you the area. If you get a minus number, just use the positive value because area can't be negative.

Exam Tip: Remember that the area of a triangle must always be a positive value. If your calculation results in a negative number, always take its absolute value.

Question 2. Show that the points A (a,b + c), B (b, c + a), C (c, a + b) are collinear.
Answer: The vertices of triangle ABC are \( A (a, b + c), B (b, c + a), C (c, a + b) \).
The area \( A \) of triangle ABC is given by the formula:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| \)
Substituting the given coordinates:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array}\right| \)
Now, we perform the column operation \( C_1 \rightarrow C_1 + C_2 \). This means we add the elements of the second column to the first column:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} a+(b+c) & b+c & 1 \\ b+(c+a) & c+a & 1 \\ c+(a+b) & a+b & 1 \end{array}\right| = \frac { 1 }{ 2 }\left|\begin{array}{lll} a+b+c & b+c & 1 \\ a+b+c & c+a & 1 \\ a+b+c & a+b & 1 \end{array}\right| \)
We can factor out \( (a+b+c) \) from the first column:
\( A = (a+b+c) \frac { 1 }{ 2 }\left|\begin{array}{lll} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1 \end{array}\right| \)
Since the first and third columns of the determinant are identical (both contain all ones), the value of the determinant is zero.
\( \implies A = (a+b+c) \frac { 1 }{ 2} \times 0 = 0 \).
Since the area of the triangle is zero, it means that the three points A, B, and C lie on the same straight line.
Hence, the points A, B, and C are collinear.
In simple words: When three points form a triangle with zero area, it means they all lie on the same straight line. We calculated the area of the triangle formed by these specific points and found it to be zero, which proves they are collinear.

Exam Tip: To prove that three points are collinear using determinants, show that the area of the triangle formed by these points is zero. This implies they lie on a single straight line.

Question 3. Find the value of k, if the area of the triangle is 4 square units and vertices are:
(i) (k, 0), (4, 0), (0, 2)
(ii) (-2, 0), (0, 4), (0, k)
Answer:
(i) The area \( A \) of the triangle with vertices \( (k, 0), (4, 0), (0, 2) \) is:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| = \frac { 1 }{ 2 }\left|\begin{array}{lll} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array}\right| \)
Expanding with the help of elements of the second column, we get:
\( A = \frac { 1 }{ 2 } [0 \times (\text{cofactor}) - 0 \times (\text{cofactor}) + 2 \times (k \times 1 - 4 \times 1)] \)
\( = \frac { 1 }{ 2 } [2(k – 4)] = k – 4 \)
Given that the area is 4 square units, we must consider both positive and negative values because the determinant can be negative, but area is always positive:
\( k – 4 = \pm 4 \)
Taking the positive sign: \( k – 4 = 4 \)
\( \implies k = 4 + 4 = 8 \).
Taking the negative sign: \( k – 4 = -4 \)
\( \implies k = -4 + 4 = 0 \).
So, the possible values for \( k \) are 0 or 8.
(ii) The area \( A \) of the triangle whose vertices are \( (-2, 0), (0, 4), (0, k) \) is:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| = \frac { 1 }{ 2 }\left|\begin{array}{ccc} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array}\right| \)
Expanding with the help of elements of the first column, we get:
\( A = \frac { 1 }{ 2 } [-2(4 \times 1 - k \times 1) - 0 \times (\text{cofactor}) + 0 \times (\text{cofactor})] \)
\( = \frac { 1 }{ 2 } [-2(4 – k)] = -(4 – k) = k - 4 \)
Given that the area is 4 square units:
\( k – 4 = \pm 4 \)
Taking the positive sign: \( k – 4 = 4 \)
\( \implies k = 4 + 4 = 8 \).
Taking the negative sign: \( k – 4 = -4 \)
\( \implies k = -4 + 4 = 0 \).
So, the possible values for \( k \) are 0 or 8.
In simple words: To find the unknown coordinate 'k' when the triangle's area is known, we use the determinant formula for area. Since area is always a positive number, the determinant's result could be positive or negative. So, we solve for 'k' using both a positive and a negative area value, giving us two possible answers for 'k'.

Exam Tip: When given the area of a triangle to find an unknown coordinate, remember to use both the positive and negative values of the given area in your calculations, as the determinant can be negative but area is not.

Question 4.
(i) Find the equation of line joining (1, 2) and (3, 6), using determinants.
(ii) Find the equation of line joining (3,1) and (9, 3), using determinants.
Answer:
(i) Let \( (x, y) \) be the third point on the line that joins \( (1, 2) \) and \( (3, 6) \).
If these three points are collinear (lie on the same line), the area of the triangle formed by them must be zero.
The area \( A \) of a triangle with vertices \( (x, y), (1, 2) \) and \( (3, 6) \) is given by:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{array}\right| \)
Since the three points are collinear, the area \( A \) is 0. So, we set the determinant to zero:
\( \left|\begin{array}{lll} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{array}\right| = 0 \)
Expanding the determinant along the first row:
\( x(2 \times 1 - 6 \times 1) - y(1 \times 1 - 3 \times 1) + 1(1 \times 6 - 3 \times 2) = 0 \)
\( x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0 \)
\( x(-4) – y(-2) + 1(0) = 0 \)
\( -4x + 2y = 0 \)
Dividing by 2, we get:
\( -2x + y = 0 \), or \( 2x - y = 0 \).
This is the required equation of the line.
(ii) Let \( (x, y) \) be the third point on the line that joins \( (3, 1) \) and \( (9, 3) \).
Since these points are collinear, the area of the triangle formed by them is zero.
The area \( A \) of a triangle with vertices \( (x, y), (3, 1) \) and \( (9, 3) \) is:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{array}\right| \)
Setting the determinant to zero:
\( \left|\begin{array}{lll} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{array}\right| = 0 \)
Expanding the determinant along the first row:
\( x(1 \times 1 - 3 \times 1) - y(3 \times 1 - 9 \times 1) + 1(3 \times 3 - 9 \times 1) = 0 \)
\( x(1 – 3) - y(3-9) + 1(9-9) = 0 \)
\( x(-2) - y(-6) + 1(0) = 0 \)
\( -2x + 6y = 0 \)
Dividing by -2, we get:
\( x - 3y = 0 \).
This is the required equation of the line.
In simple words: To find the equation of a line using determinants, we pick any point (x, y) on that line. Since three points on a line don't make a triangle, the area of the "triangle" formed by these three points (your two given points plus (x, y)) must be zero. Setting the determinant for area to zero helps you find the line's equation.

Exam Tip: When using the determinant method to find the equation of a line, always remember that collinear points result in a zero area, which simplifies the determinant expression into the linear equation.

Question 5. If the area of a triangle with vertices (2, – 6), (5, 4) and (k, 4) is 35 sq. units, then k is
(a) 12
(b) -2
(c) -12, -2
(d) 12, -2
Answer: (d) 12, -2
The area \( A \) of the triangle with the given vertices \( (2, -6), (5, 4) \) and \( (k, 4) \) is given by:
\( A = \frac { 1 }{ 2 }\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| = \frac { 1 }{ 2 }\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right| \)
Expanding the determinant along the first row:
\( A = \frac { 1 }{ 2 }[2(4 \times 1 - 4 \times 1) - (-6)(5 \times 1 - k \times 1) + 1(5 \times 4 - k \times 4)] \)
\( = \frac { 1 }{ 2 }[2(4-4) + 6(5 -k) + 1(20 – 4k) ] \)
\( = \frac { 1 }{ 2 }[2(0) + 30 – 6k + 20 – 4k ] \)
\( = \frac { 1 }{ 2 }[50 – 10k] \).
Given that the area is 35 sq. units, we must account for both positive and negative values of the determinant, as area is absolute:
\( \frac { 1 }{ 2 } [50 – 10k] = \pm 35 \)
Multiply both sides by 2:
\( 50 – 10k = \pm 70 \)
Case 1: Taking the positive sign
\( 50 – 10k = 70 \)
\( -10k = 70 - 50 \)
\( -10k = 20 \)
\( k = \frac { 20 }{ -10 } \)
\( \implies k = -2 \).
Case 2: Taking the negative sign
\( 50 – 10k = -70 \)
\( -10k = -70 - 50 \)
\( -10k = -120 \)
\( k = \frac { -120 }{ -10 } \)
\( \implies k = 12 \).
Hence, the possible values for \( k \) are 12 and -2.
In simple words: We know the triangle's area and its corner points, with one point having an unknown 'k' value. We use a math formula called a determinant to set up an equation for the area. Since area is always positive, the calculated part can be either positive or negative, so we solve for 'k' in both situations. This gives us two possible 'k' values.

Exam Tip: For MCQ questions involving area and an unknown coordinate, always remember to consider both positive and negative values of the given area in your determinant calculation, as this often leads to multiple valid solutions for the unknown variable.