Samacheer Kalvi Class 11 Business Maths Solutions Chapter 3 Analytical Geometry Exercise 3.2

 

Question 1. Find the angle between the lines whose slopes are \( \frac{1}{2} \) and 3.
Answer: Let \( m_1 = \frac{1}{2} \) and \( m_2 = 3 \) be the slopes of the two lines. The formula to find the angle \( \theta \) between two lines is given by \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \). This formula helps us determine the acute angle between intersecting lines using their gradients.
Now, we substitute the given slopes into the formula:
\( \tan \theta = \left| \frac{\frac{1}{2} - 3}{1 + \frac{1}{2} \times 3} \right| \)
\( \tan \theta = \left| \frac{\frac{1-6}{2}}{1 + \frac{3}{2}} \right| \)
\( \tan \theta = \left| \frac{-\frac{5}{2}}{\frac{2+3}{2}} \right| \)
\( \tan \theta = \left| \frac{-\frac{5}{2}}{\frac{5}{2}} \right| \)
\( \tan \theta = |-1| \)
\( \tan \theta = 1 \)
We know that \( \tan 45^\circ = 1 \).
So, \( \theta = 45^\circ \).
In simple words: We use a special formula that connects the slopes of two lines to the angle between them. By putting the given slopes into this formula, we calculated that the angle between the lines is 45 degrees.

🎯 Exam Tip: Always remember to use the absolute value in the tan formula for the angle between lines, as the angle is usually considered acute (between 0 and 90 degrees).

 

Question 2. Find the distance of the point (4, 1) from the line 3x - 4y + 12 = 0.
Answer: We need to find the perpendicular distance from the point \( (x_1, y_1) = (4, 1) \) to the line \( ax + by + c = 0 \), which is \( 3x - 4y + 12 = 0 \). The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is \( d = \left| \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \right| \). This formula is very useful for calculating the shortest distance between a point and a line.
Here, \( a = 3 \), \( b = -4 \), \( c = 12 \), and the point is \( (x_1, y_1) = (4, 1) \).
Substitute these values into the formula:
\( d = \left| \frac{3(4) - 4(1) + 12}{\sqrt{3^2 + (-4)^2}} \right| \)
\( d = \left| \frac{12 - 4 + 12}{\sqrt{9 + 16}} \right| \)
\( d = \left| \frac{20}{\sqrt{25}} \right| \)
\( d = \left| \frac{20}{5} \right| \)
\( d = 4 \)
The distance of the point (4, 1) from the line 3x - 4y + 12 = 0 is 4 units.
In simple words: To find how far a point is from a straight line, we use a special formula. We put the numbers from our point and line into this formula, and after doing some calculations, we find the distance is 4 units.

🎯 Exam Tip: Remember to use the absolute value at the end of the distance formula, as distance is always a positive quantity.

 

Question 3. Show that the straight lines x + y - 4 = 0, 3x + 2 = 0 and 3x - 3y + 16 = 0 are concurrent.
Answer: For three lines \( a_1 x + b_1 y + c_1 = 0 \), \( a_2 x + b_2 y + c_2 = 0 \), and \( a_3 x + b_3 y + c_3 = 0 \) to be concurrent (meaning they all meet at one single point), the determinant of their coefficients must be zero. This condition ensures that a unique intersection point exists for all three lines.
The given lines are:
1. \( x + y - 4 = 0 \implies a_1 = 1, b_1 = 1, c_1 = -4 \)
2. \( 3x + 0y + 2 = 0 \implies a_2 = 3, b_2 = 0, c_2 = 2 \)
3. \( 3x - 3y + 16 = 0 \implies a_3 = 3, b_3 = -3, c_3 = 16 \)
Now, we calculate the determinant:
\[ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = \begin{vmatrix} 1 & 1 & -4 \\ 3 & 0 & 2 \\ 3 & -3 & 16 \end{vmatrix} \] Expanding the determinant along the first row:
\( = 1(0 \times 16 - 2 \times (-3)) - 1(3 \times 16 - 2 \times 3) + (-4)(3 \times (-3) - 0 \times 3) \)
\( = 1(0 + 6) - 1(48 - 6) - 4(-9 - 0) \)
\( = 1(6) - 1(42) - 4(-9) \)
\( = 6 - 42 + 36 \)
\( = 42 - 42 \)
\( = 0 \)
Since the determinant is 0, the given lines are concurrent.
In simple words: We check if three lines meet at one point by calculating a special number called a determinant from their equations. If this number turns out to be zero, it means all three lines cross at the same spot. In this problem, the determinant was zero, so the lines are concurrent.

🎯 Exam Tip: Remember the condition for concurrency of three lines using the determinant of their coefficients. Carefully compute the determinant to avoid calculation errors.

 

Question 4. Find the value of 'a' for which the straight lines 3x + 4y = 13; 2x - 7y = -1 and ax - y - 14 = 0 are concurrent.
Answer: For the three given lines to be concurrent, the determinant of their coefficients must be equal to zero. This is a key principle in analytical geometry to verify if multiple lines share a common intersection point.
First, rewrite the equations in the standard form \( ax + by + c = 0 \):
1. \( 3x + 4y - 13 = 0 \implies a_1 = 3, b_1 = 4, c_1 = -13 \)
2. \( 2x - 7y + 1 = 0 \implies a_2 = 2, b_2 = -7, c_2 = 1 \)
3. \( ax - y - 14 = 0 \implies a_3 = a, b_3 = -1, c_3 = -14 \)
Set up the determinant and set it equal to zero:
\[ \begin{vmatrix} 3 & 4 & -13 \\ 2 & -7 & 1 \\ a & -1 & -14 \end{vmatrix} = 0 \] Expand the determinant:
\( 3((-7) \times (-14) - 1 \times (-1)) - 4(2 \times (-14) - 1 \times a) + (-13)(2 \times (-1) - (-7) \times a) = 0 \)
\( 3(98 + 1) - 4(-28 - a) - 13(-2 + 7a) = 0 \)
\( 3(99) + 112 + 4a + 26 - 91a = 0 \)
\( 297 + 112 + 26 + 4a - 91a = 0 \)
\( 435 - 87a = 0 \)
Now, solve for \( a \):
\( -87a = -435 \)

\( \implies a = \frac{-435}{-87} \)

\( \implies a = 5 \)
So, the value of 'a' for which the lines are concurrent is 5.
In simple words: For three lines to meet at one point, a special number made from their equation parts must be zero. We set up this calculation, which involves the unknown 'a', and then solve it like a puzzle to find that 'a' must be 5.

🎯 Exam Tip: Double-check your arithmetic when expanding the determinant, as sign errors are common and can lead to an incorrect value of 'a'.

 

Question 5. A manufacturer produces 80 TV sets at a cost of Rs. 2,20,000 and 125 TV sets at a cost of Rs. 2,87,500. Assuming the cost curve to be linear, find the linear expression. Also, estimate the cost of 95 TV sets.
Answer: Let \( x \) represent the number of TV sets and \( y \) represent the cost. A linear cost curve means the relationship between the number of TV sets and their cost can be shown as a straight line. We can find the equation of this line using the two given points: \( (x_1, y_1) = (80, 220000) \) and \( (x_2, y_2) = (125, 287500) \). Understanding the linear relationship helps in predicting costs for different production levels.
We can use the two-point form of a straight line equation: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \).
First, let's represent the given information in a table:

Substitute the values into the formula:
\( \frac{y - 220000}{287500 - 220000} = \frac{x - 80}{125 - 80} \)
\( \frac{y - 220000}{67500} = \frac{x - 80}{45} \)
Now, simplify the fraction on the left side: \( \frac{67500}{45} = 1500 \).
\( \frac{y - 220000}{1500} = \frac{x - 80}{1} \)
Multiply both sides by 1500:
\( y - 220000 = 1500(x - 80) \)
\( y - 220000 = 1500x - 1500 \times 80 \)
\( y - 220000 = 1500x - 120000 \)
Add 220000 to both sides to solve for \( y \):
\( y = 1500x - 120000 + 220000 \)
\( y = 1500x + 100000 \)
This is the linear expression (cost function).

Next, we estimate the cost of 95 TV sets. We use the linear expression we just found by setting \( x = 95 \).
\( y = 1500 \times 95 + 100000 \)
\( y = 142500 + 100000 \)
\( y = 242500 \)
Therefore, the estimated cost of 95 TV sets is Rs. 2,42,500.
In simple words: We are given costs for making two different numbers of TV sets. Since the cost goes up in a straight line, we found a simple math rule (an equation) that tells us the cost for any number of TVs. Using this rule, we figured out that making 95 TV sets would cost Rs. 2,42,500.

🎯 Exam Tip: When dealing with linear equations, clearly identify your two points (x1, y1) and (x2, y2) and use the two-point form of the line equation. Always double-check your arithmetic, especially when working with large numbers.