GSEB Class 12 Maths Solutions Chapter 4 નિશ્ચાયક Exercise 4.3

Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.3

 

Question 1. નીચે આપેલાં શિરોબિંદુવાળા ત્રિકોણનું ક્ષેત્રફળ શોધો:
(i) (1, 0), (6, 0), (4, 3)
Answer: Let A(1, 0), B(6, 0), and C(4, 3) be the vertices of triangle ABC. The area of triangle ABC is calculated using the formula \( \frac{1}{2}|D| \), where D is the determinant of the matrix formed by the coordinates.
Here, D is:

1	0	1
6	0	1
4	3	1

We compute the determinant:
\( D = 1(0-3) - 0(6-4) + 1(18-0) \)
\( = -3 + 18 \)
\( = 15 \) So, the area of triangle ABC is \( \frac{1}{2}|15| = \frac{15}{2} \) square units.
In simple words: We find the area of the triangle by using a special calculation called a determinant, which uses the coordinates of its corners.

🎯 Exam Tip: Remember the formula for the area of a triangle using determinants: \( \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \). Ensure correct sign conventions during expansion.

 

Question 1. (Continued)
(ii) (2, 7), (1, 1), (10, 8)
Answer: Let A(2, 7), B(1, 1), and C(10, 8) be the vertices of triangle ABC. The area of triangle ABC is found using the formula \( \frac{1}{2}|D| \), where D is the determinant.
Here, D is:

2	7	1
1	1	1
10	8	1

We compute the determinant:
\( D = 2(1-8) - 7(1-10) + 1(8-10) \)
\( = 2(-7) - 7(-9) + 1(-2) \)
\( = -14 + 63 - 2 \)
\( = 47 \) Therefore, the area of triangle ABC is \( \frac{1}{2}|47| = \frac{47}{2} \) square units.
In simple words: This calculates the area of a triangle using its three corner points by performing a specific determinant operation.

🎯 Exam Tip: When calculating determinants, be careful with signs. A common mistake is mismanaging the negative sign for the middle term in a 3x3 determinant expansion.

 

Question 1. (Continued)
(iii) (-2,-3), (3, 2), (-1, -8)
Answer: Let A(-2, -3), B(3, 2), and C(-1, -8) be the vertices of triangle ABC. The area of triangle ABC is determined by \( \frac{1}{2}|D| \), where D is the determinant.
Here, D is:

-2	-3	1
3	2	1
-1	-8	1

We compute the determinant:
\( D = -2(2+8) - (-3)(3+1) + 1(-24+2) \)
\( = -2(10) + 3(4) + 1(-22) \)
\( = -20 + 12 - 22 \)
\( = -30 \) Since the area must be positive, the area of triangle ABC is \( \frac{1}{2}|-30| = \frac{30}{2} = 15 \) square units.
In simple words: We find the area of a triangle using the determinant of its vertex coordinates, taking the absolute value to ensure a positive area.

🎯 Exam Tip: The area of a triangle cannot be negative. If the determinant calculation yields a negative value, take its absolute value before dividing by two to get the actual area.

 

Question 2. સાબિત કરો કે બિંદુઓ A(a, b + c), B(b, c + a) du C(c, a + b) સમરેખ છે.
Answer: To prove that the points A(a, b + c), B(b, c + a), and C(c, a + b) are collinear (lie on the same line), we need to show that the area of the triangle formed by these points is zero. We calculate the determinant D: \[ D = \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix} \] Apply the column operation \( C_1 \to C_1 + C_2 \): \[ D = \begin{vmatrix} a+b+c & b+c & 1 \\ b+c+a & c+a & 1 \\ c+a+b & a+b & 1 \end{vmatrix} \] Now, take out \( (a+b+c) \) as a common factor from the first column: \[ D = (a+b+c) \begin{vmatrix} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1 \end{vmatrix} \] In this new determinant, the first column \( (C_1) \) and the third column \( (C_3) \) are identical. If two columns (or rows) of a determinant are identical, its value is zero. So, \( D = (a+b+c)(0) \)
\( = 0 \) Since the determinant D is 0, the area of the triangle formed by these points is \( \frac{1}{2}|D| = 0 \). Therefore, the points are collinear.
In simple words: When three points lie on the same straight line, they cannot form a triangle, meaning the area of the "triangle" they would make is zero. We prove this by calculating a specific math value called a determinant, and if it's zero, the points are on one line.

🎯 Exam Tip: For proving collinearity, showing that the area of the triangle formed by the points is zero using the determinant method is the most direct and accepted approach. Remember that if any two rows or columns of a determinant are identical, its value is zero.

 

Question 3. જો (i) (k, 0) (4, 0), (0, 2), (ii) (-2, 0) (0, 4), (0, k) શિરોબિંદુવાળા ત્રિકોણનું ક્ષેત્રફળ 4 ચોરસ એકમ હોય, તો k નું મૂલ્ય શોધો.
Answer: (i) Given the vertices (k, 0), (4, 0), and (0, 2). The area of the triangle is 4 square units. First, calculate the determinant D: \[ D = \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} \] Expand the determinant along the second column (because it has two zeros):
\( D = -0 \begin{vmatrix} 4 & 1 \\ 0 & 1 \end{vmatrix} + 0 \begin{vmatrix} k & 1 \\ 0 & 1 \end{vmatrix} - 2 \begin{vmatrix} k & 1 \\ 4 & 1 \end{vmatrix} \)
\( = -2(k \cdot 1 - 4 \cdot 1) \)
\( = -2(k - 4) \)
\( = -2k + 8 \) The area of the triangle is \( \frac{1}{2}|D| \). We are given that the area is 4. So, \( 4 = \frac{1}{2}|-2k + 8| \)
\( 8 = |-2k + 8| \) This means \( -2k + 8 = 8 \) or \( -2k + 8 = -8 \). Case 1: \( -2k + 8 = 8 \)
\( -2k = 0 \)
\( k = 0 \) Case 2: \( -2k + 8 = -8 \)
\( -2k = -16 \)
\( k = 8 \) Thus, the value of k is 0 or 8.
In simple words: We are given the area of a triangle and two of its corner points, with one coordinate of the third point missing. We use the area formula with determinants to set up an equation and solve for the missing 'k' value, finding two possible answers.

🎯 Exam Tip: When solving equations involving absolute values, remember to consider both the positive and negative possibilities for the expression inside the absolute value, as \( |x| = a \) means \( x = a \) or \( x = -a \).

 

Question 3. (Continued)
(ii) Given the vertices (-2, 0), (0, 4), and (0, k). The area of the triangle is 4 square units.
Answer: First, calculate the determinant D: \[ D = \begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} \] Expand the determinant along the first column:
\( D = -2(4 \cdot 1 - k \cdot 1) - 0 \begin{vmatrix} 0 & 1 \\ k & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 4 & 1 \end{vmatrix} \)
\( = -2(4 - k) \)
\( = -8 + 2k \) The area of the triangle is \( \frac{1}{2}|D| \). We are given that the area is 4. So, \( 4 = \frac{1}{2}|-8 + 2k| \)
\( 8 = |-8 + 2k| \) This means \( -8 + 2k = 8 \) or \( -8 + 2k = -8 \). Case 1: \( -8 + 2k = 8 \)
\( 2k = 16 \)
\( k = 8 \) Case 2: \( -8 + 2k = -8 \)
\( 2k = 0 \)
\( k = 0 \) Thus, the value of k is 8 or 0.
In simple words: Similar to the previous part, we use the known triangle area and two given points, plus one point with a missing coordinate, to form a determinant. Solving the resulting equation gives us the possible values for 'k'.

🎯 Exam Tip: Always double-check your determinant expansion, especially when dealing with negative coordinates or variables. Organize your steps to avoid arithmetic errors.

 

Question 4. (i) નિશ્ચાયકનો ઉપયોગ કરી (1, 2) અને (3, 6) ને જોડતી રેખાનું સમીકરણ શોધો.
Answer: To find the equation of the line passing through points A(1, 2) and B(3, 6) using determinants, we can consider a general point P(x, y) on the line. Since A, B, and P are collinear, the area of the triangle formed by them must be zero. So, the determinant is: \[ \begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0 \] Expand the determinant:
\( x(2 \cdot 1 - 6 \cdot 1) - y(1 \cdot 1 - 3 \cdot 1) + 1(1 \cdot 6 - 3 \cdot 2) = 0 \)
\( x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0 \)
\( x(-4) - y(-2) + 1(0) = 0 \)
\( -4x + 2y = 0 \) Divide by -2:
\( 2x - y = 0 \) Therefore, the equation of the line is \( y = 2x \).
In simple words: To find the equation of a straight line using determinants, we imagine any point (x,y) on the line. Since this point and the two given points are all on the same line, they can't form a triangle, so the area calculation (determinant) for these three points must be zero. Solving this gives the line's equation.

🎯 Exam Tip: When forming the determinant for the equation of a line, the first row should always be (x, y, 1), and the subsequent rows are the coordinates of the given points, each followed by a 1. Set the determinant equal to zero.

 

Question 4. (Continued)
(ii) નિશ્ચાયકનો ઉપયોગ કરી (3, 1) અને (9, 3) ને જોડતી રેખાનું સમીકરણ શોધો.
Answer: To find the equation of the line passing through points A(3, 1) and B(9, 3) using determinants, we consider a general point P(x, y) on the line. Since A, B, and P are collinear, the area of the triangle formed by them must be zero. So, the determinant is: \[ \begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0 \] Expand the determinant:
\( x(1 \cdot 1 - 3 \cdot 1) - y(3 \cdot 1 - 9 \cdot 1) + 1(3 \cdot 3 - 9 \cdot 1) = 0 \)
\( x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0 \)
\( x(-2) - y(-6) + 1(0) = 0 \)
\( -2x + 6y = 0 \) Divide by -2:
\( x - 3y = 0 \) Therefore, the equation of the line is \( x = 3y \).
In simple words: We find the straight line equation using determinants. By assuming a general point (x,y) on the line, and knowing that this point along with the two given points form no triangle (zero area), we solve the determinant to get the line's formula.

🎯 Exam Tip: After expanding the determinant and simplifying the equation, always try to express it in its simplest form (e.g., \( Ax + By + C = 0 \) or \( y = mx + c \)) by dividing by common factors if possible.

 

Question 5. જો (2, -6), (5, 4) અને (k, 4) શિરોબિંદુવાળા ત્રિકોણનું ક્ષેત્રફળ 35 ચોરસ એકમ હોય, તો k નું મૂલ્ય _______.
(A) 12
(B) -2
(C) -12, -2
(D) 12, -2
Answer: Given the vertices (2, -6), (5, 4), and (k, 4). The area of the triangle is 35 square units. First, calculate the determinant D: \[ D = \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} \] Expand the determinant:
\( D = 2(4 \cdot 1 - 4 \cdot 1) - (-6)(5 \cdot 1 - k \cdot 1) + 1(5 \cdot 4 - k \cdot 4) \)
\( = 2(4 - 4) + 6(5 - k) + 1(20 - 4k) \)
\( = 2(0) + 30 - 6k + 20 - 4k \)
\( = 0 + 30 - 6k + 20 - 4k \)
\( = 50 - 10k \) The area of the triangle is \( \frac{1}{2}|D| \). We are given that the area is 35. So, \( 35 = \frac{1}{2}|50 - 10k| \)
\( 70 = |50 - 10k| \) This means \( 50 - 10k = 70 \) or \( 50 - 10k = -70 \). Case 1: \( 50 - 10k = 70 \)
\( -10k = 20 \)
\( k = -2 \) Case 2: \( 50 - 10k = -70 \)
\( -10k = -120 \)
\( k = 12 \) Thus, the value of k is 12 or -2.
Answer: (D) 12, -2
In simple words: We use the given area of a triangle and the coordinates of its corners, where one coordinate is unknown 'k'. By setting up a determinant equation and solving it, we find two possible values for 'k' because the absolute value of the determinant is used for area.

🎯 Exam Tip: For MCQ questions, after finding the possible values of k, always check them against the given options to ensure you select the correct choice. Remember that a determinant can be positive or negative, but area is always positive.