Samacheer Kalvi Class 11 Business Maths Solutions Chapter 3 Analytical Geometry Exercise 3.3

 

Question 1. If the equation \( ax^2 + 5xy - 6y^2 + 12x + 5y + c = 0 \) represents a pair of straight lines, find a and c.
Answer: To find 'a' and 'c', we first compare the given equation \( ax^2 + 5xy - 6y^2 + 12x + 5y + c = 0 \) with the general equation of a pair of straight lines, \( Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \). This comparison gives us: \( A = a \), \( 2H = 5 \implies H = \frac{5}{2} \), \( B = -6 \), \( 2G = 12 \implies G = 6 \), \( 2F = 5 \implies F = \frac{5}{2} \), and \( C = c \). For the straight lines to be perpendicular, a condition is that \( A + B = 0 \). So, \( a + (-6) = 0 \).
\( \implies a = 6 \). Next, we find 'c'. For the given equation to represent a pair of straight lines, a special condition must be met: the determinant of the coefficient matrix must be zero. This helps us ensure the equation forms actual lines. The determinant is: \[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \] Substituting the values \( a=6, h=\frac{5}{2}, b=-6, g=6, f=\frac{5}{2} \):

To solve the determinant more easily, we can use a row operation: \( R_1 \rightarrow R_1 - R_3 \). This changes the first row to \( (6-6) \), \( (\frac{5}{2} - \frac{5}{2}) \), \( (6-c) \), which is \( (0, 0, 6-c) \). The determinant becomes:Expanding this determinant along the first row, we only need to consider the \( (6-c) \) term: \( (6-c) \left[ \left(\frac{5}{2}\right) \left(\frac{5}{2}\right) - (6)(-6) \right] = 0 \)
\( \implies (6-c) \left[ \frac{25}{4} + 36 \right] = 0 \)
\( \implies (6-c) \left[ \frac{25 + 144}{4} \right] = 0 \)
\( \implies (6-c) \left[ \frac{169}{4} \right] = 0 \) Since \( \frac{169}{4} \) is not zero, the term \( (6-c) \) must be zero.
\( \implies 6 - c = 0 \)
\( \implies c = 6 \) So, the values are \( a = 6 \) and \( c = 6 \).In simple words: We compare the given equation with the standard form for straight lines to find values like 'a', 'h', 'b', 'g', 'f', and 'c'. For perpendicular lines, 'a+b' must be zero, which helps us find 'a'. Then, for the equation to represent lines, a special calculation called a determinant must be zero. Solving this determinant, often by simplifying the matrix, helps us find 'c'. Both 'a' and 'c' turn out to be 6.

🎯 Exam Tip: Remember to clearly state the general equation of a pair of straight lines and the conditions for specific properties, like perpendicularity or representing lines, as these are key steps in scoring full marks.

 

Question 2. Show that the equation \( 12x^2 - 10xy + 2y^2 + 14x - 5y + 2 = 0 \) represents a pair of straight lines and also find the separate equations of the straight lines.
Answer: To show that the equation represents a pair of straight lines, we compare the given equation \( 12x^2 - 10xy + 2y^2 + 14x - 5y + 2 = 0 \) with the general equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \). From this comparison, we get the coefficients: \( a = 12 \) \( 2h = -10 \implies h = -5 \) \( b = 2 \) \( 2g = 14 \implies g = 7 \) \( 2f = -5 \implies f = -\frac{5}{2} \) \( c = 2 \) For the equation to represent a pair of straight lines, the determinant of the coefficient matrix must be equal to zero. This is a key condition for such equations. The determinant is: \[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \] Substituting the values we found:

Now we calculate the determinant: \( 12 \left[ (2)(2) - \left(-\frac{5}{2}\right)\left(-\frac{5}{2}\right) \right] - (-5) \left[ (-5)(2) - 7\left(-\frac{5}{2}\right) \right] + 7 \left[ (-5)\left(-\frac{5}{2}\right) - 7(2) \right] = 0 \)
\( \implies 12 \left[ 4 - \frac{25}{4} \right] + 5 \left[ -10 + \frac{35}{2} \right] + 7 \left[ \frac{25}{2} - 14 \right] = 0 \)
\( \implies 12 \left[ \frac{16 - 25}{4} \right] + 5 \left[ \frac{-20 + 35}{2} \right] + 7 \left[ \frac{25 - 28}{2} \right] = 0 \)
\( \implies 12 \left[ -\frac{9}{4} \right] + 5 \left[ \frac{15}{2} \right] + 7 \left[ -\frac{3}{2} \right] = 0 \)
\( \implies -27 + \frac{75}{2} - \frac{21}{2} = 0 \)
\( \implies -27 + \frac{54}{2} = 0 \)
\( \implies -27 + 27 = 0 \)
\( \implies 0 = 0 \) Since the determinant is 0, the given equation indeed represents a pair of straight lines. Next, we find the separate equations of these lines. Consider the homogeneous part of the equation: \( 12x^2 - 10xy + 2y^2 \). We can factor this part: \( 2(6x^2 - 5xy + y^2) = 2(3x - y)(2x - y) = (6x - 2y)(2x - y) \). Let the two separate lines be \( (6x - 2y + l) = 0 \) and \( (2x - y + m) = 0 \). Multiplying these and comparing with the original equation: \( (6x - 2y + l)(2x - y + m) = 12x^2 - 6xy + 6xm - 4xy + 2y^2 - 2ym + 2xl - yl + lm \) \( = 12x^2 - 10xy + 2y^2 + (6m + 2l)x + (-2m - l)y + lm \) Comparing the coefficients with the original equation \( 12x^2 - 10xy + 2y^2 + 14x - 5y + 2 = 0 \): Coefficient of x: \( 6m + 2l = 14 \implies 3m + l = 7 \) (Equation 2) Coefficient of y: \( -2m - l = -5 \implies 2m + l = 5 \) (Equation 3) Constant term: \( lm = 2 \) From Equation 2, we can express \( l \) as: \( l = 7 - 3m \). Substitute this expression for \( l \) into Equation 3: \( 2m + (7 - 3m) = 5 \) \( \implies 7 - m = 5 \)
\( \implies m = 2 \) Now, substitute \( m = 2 \) back into the expression for \( l \): \( l = 7 - 3(2) \) \( l = 7 - 6 \)
\( \implies l = 1 \) Let's check these values with the constant term condition: \( lm = (1)(2) = 2 \), which matches the original equation's constant term. So, the separate equations of the straight lines are: \( 6x - 2y + 1 = 0 \) \( 2x - y + 2 = 0 \)In simple words: First, we prove that the equation forms a pair of straight lines by checking if a special determinant calculation is zero. After comparing coefficients, we set up this determinant and find that it is indeed zero. This means the equation represents two lines. Next, to find the equations of these two separate lines, we break down the first part of the original equation (the terms with x squared, y squared, and xy) into two factors. Then, we add general constant terms ('l' and 'm') to these factors. By multiplying these two new line equations and comparing them with the full original equation, we can find 'l' and 'm'. With 'l' as 1 and 'm' as 2, we get the two individual line equations.

🎯 Exam Tip: Always remember that showing the determinant is zero is crucial for proving a general second-degree equation represents a pair of straight lines. For factorizing, focus on the homogeneous part first.

 

Question 3. Show that the pair of straight lines \( 4x^2 + 12xy + 9y^2 - 6x - 9y + 2 = 0 \) represents two parallel straight lines and also find the separate equations of the straight lines.
Answer: To show the equation represents two parallel straight lines, we first compare the given equation \( 4x^2 + 12xy + 9y^2 - 6x - 9y + 2 = 0 \) with the general form \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \). From this comparison, we get the coefficients: \( a = 4 \) \( 2h = 12 \implies h = 6 \) \( b = 9 \) \( 2g = -6 \implies g = -3 \) \( 2f = -9 \implies f = -\frac{9}{2} \) \( c = 2 \) For the equation to represent parallel straight lines, two important conditions must be met: 1. The value of \( h^2 - ab \) must be zero. This tells us the lines are parallel or coincident. 2. The determinant of the coefficient matrix \( \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \) must also be zero. This ensures the equation truly represents a pair of lines. Let's check the first condition: \( h^2 - ab = (6)^2 - (4)(9) = 36 - 36 = 0 \). Since \( h^2 - ab = 0 \), this indicates that if the equation represents a pair of straight lines, they must indeed be parallel. Now, let's verify the second condition, the determinant:

Calculating the determinant: \( 4 \left[ (9)(2) - \left(-\frac{9}{2}\right)\left(-\frac{9}{2}\right) \right] - 6 \left[ (6)(2) - (-3)\left(-\frac{9}{2}\right) \right] + (-3) \left[ (6)\left(-\frac{9}{2}\right) - (-3)(9) \right] \)
\( \implies 4 \left[ 18 - \frac{81}{4} \right] - 6 \left[ 12 - \frac{27}{2} \right] - 3 \left[ -27 + 27 \right] \)
\( \implies 4 \left[ \frac{72 - 81}{4} \right] - 6 \left[ \frac{24 - 27}{2} \right] - 3 [0] \)
\( \implies 4 \left[ -\frac{9}{4} \right] - 6 \left[ -\frac{3}{2} \right] + 0 \)
\( \implies -9 - (-9) = -9 + 9 = 0 \)
\( \implies 0 = 0 \) Since both conditions (\( h^2 - ab = 0 \) and determinant = 0) are met, the equation definitely represents a pair of parallel straight lines. Now, we find the separate equations for these lines. Consider the second-degree terms: \( 4x^2 + 12xy + 9y^2 \). This expression is a perfect square: \( (2x)^2 + 2(2x)(3y) + (3y)^2 = (2x + 3y)^2 \). So the given equation can be rewritten as \( (2x + 3y)^2 - 6x - 9y + 2 = 0 \). We can factor out -3 from the terms \( -6x - 9y \): \( -3(2x + 3y) \). So the equation becomes: \( (2x + 3y)^2 - 3(2x + 3y) + 2 = 0 \). Let's simplify this by substituting \( t = 2x + 3y \). The equation becomes: \( t^2 - 3t + 2 = 0 \) This is a quadratic equation in \( t \). We can factor it: \( (t - 1)(t - 2) = 0 \) Now, substitute back \( t = 2x + 3y \): \( (2x + 3y - 1)(2x + 3y - 2) = 0 \) Therefore, the separate equations of the parallel straight lines are: \( 2x + 3y - 1 = 0 \) \( 2x + 3y - 2 = 0 \)In simple words: We first compare the given equation with the general form to find coefficients 'a', 'h', 'b', 'g', 'f', and 'c'. To show the lines are parallel, we check if \( h^2 - ab = 0 \) and if the determinant of the coefficients is zero. Both conditions are met, so the lines are indeed parallel. To find the actual equations, we notice that the terms with \( x^2, xy, y^2 \) form a perfect square, \( (2x + 3y)^2 \). We then substitute \( t = 2x + 3y \) into the full equation, which simplifies it to a basic quadratic equation in 't'. Solving this quadratic equation and putting back \( 2x + 3y \) for 't' gives us the two separate parallel line equations.

🎯 Exam Tip: When proving parallel lines, ensure you check both the \( h^2 - ab = 0 \) condition and the determinant condition. When factorizing, spotting perfect squares like \( (Ax+By)^2 \) can greatly simplify the process.

 

Question 4. Find the angle between the pair of straight lines \( 3x^2 - 5xy - 2y^2 + 17x + y + 10 = 0 \).
Answer: To find the angle between the lines, we first compare the given equation \( 3x^2 - 5xy - 2y^2 + 17x + y + 10 = 0 \) with the general equation of a pair of straight lines \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \). From this comparison, we get the coefficients: \( a = 3 \) \( 2h = -5 \implies h = -\frac{5}{2} \) \( b = -2 \) \( 2g = 17 \implies g = \frac{17}{2} \) \( 2f = 1 \implies f = \frac{1}{2} \) \( c = 10 \) The formula for the angle \( \theta \) between a pair of straight lines is given by: \[ \tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right| \] First, we need to calculate the value of \( h^2 - ab \): \( h^2 - ab = \left(-\frac{5}{2}\right)^2 - (3)(-2) \)
\( \implies h^2 - ab = \frac{25}{4} + 6 \)
\( \implies h^2 - ab = \frac{25 + 24}{4} = \frac{49}{4} \) Now, we substitute these values into the formula for \( \tan \theta \): \( \tan \theta = \left| \frac{2\sqrt{\frac{49}{4}}}{3 + (-2)} \right| \)
\( \implies \tan \theta = \left| \frac{2 \times \frac{7}{2}}{1} \right| \)
\( \implies \tan \theta = \left| \frac{7}{1} \right| \)
\( \implies \tan \theta = 7 \) Therefore, the angle \( \theta \) between the pair of straight lines is \( \tan^{-1}(7) \).In simple words: To find the angle between the lines, we first compare the given equation to the general form of a pair of straight lines to get the values for 'a', 'h', and 'b'. Then, we use a specific formula involving these values, \( \tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right| \). We calculate \( h^2 - ab \) and \( a+b \) and substitute them into the formula. After simplifying, we get the value of \( \tan \theta \). The angle \( \theta \) itself is then found by taking the inverse tangent of that value.

🎯 Exam Tip: Always remember the formula for the angle between a pair of straight lines. Ensure you correctly identify 'a', 'h', and 'b' from the general equation and carefully calculate \( h^2 - ab \) to avoid errors in the square root part.