Samacheer Kalvi Class 11 Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Exercise 6.3

 

Question 1. Show that the lines \( 3x + 2y + 9 = 0 \) and \( 12x + 8y - 15 = 0 \) are parallel lines.
Answer: The given equations for the lines are:
\( 3x + 2y + 9 = 0 \quad \text{(1)} \)
\( 12x + 8y - 15 = 0 \quad \text{(2)} \)
To check if lines are parallel, we find their slopes. The slope \( m \) of a line in the form \( Ax + By + C = 0 \) is given by \( m = -\frac{A}{B} \).
For line (1): \( m_1 = - \frac{3}{2} \)
For line (2): \( m_2 = - \frac{12}{8} \)
We can simplify \( m_2 \): \( m_2 = - \frac{3 \times 4}{2 \times 4} = - \frac{3}{2} \).
Since both slopes are equal, \( m_1 = m_2 = - \frac{3}{2} \), the given lines are parallel. This is because parallel lines always have the same slope.
In simple words: We found the slope for each line. Since both lines have the exact same slope, it means they run alongside each other and never cross, so they are parallel.

🎯 Exam Tip: Remember that for lines in the form \( Ax + By + C = 0 \), the slope is \( -A/B \). Always simplify the slopes to their lowest terms to easily compare them.

 

Question 2. Find the equation of the straight line parallel to \( 5x - 4y + 3 = 0 \) and having X-intercept 3.
Answer: Any line parallel to \( 5x - 4y + 3 = 0 \) will have the same slope, so its equation can be written as:
\( 5x - 4y + k = 0 \quad \text{(1)} \)
We are given that the x-intercept of this new line is 3. An x-intercept occurs when the y-coordinate is 0.
So, we substitute \( y = 0 \) into equation (1):
\( 5x - 4(0) + k = 0 \)
\( 5x + k = 0 \)
\( 5x = -k \)
\( x = - \frac{k}{5} \)
Since the x-intercept is 3, we have:
\( - \frac{k}{5} = 3 \)
Multiply both sides by -5 to find \( k \):
\( k = -15 \)
Now, substitute the value of \( k \) back into the equation (1) to get the required line's equation:
\( 5x - 4y - 15 = 0 \). This line satisfies both conditions given in the problem.
In simple words: We first wrote down what a parallel line would generally look like. Then, we used the information about where it crosses the x-axis (the x-intercept) to find the missing number in our line's equation.

🎯 Exam Tip: For lines parallel to \( Ax + By + C = 0 \), their equation is generally \( Ax + By + k = 0 \). To find the x-intercept, set \( y = 0 \), and to find the y-intercept, set \( x = 0 \).

 

Question 3. Find the distance between the line \( 4x + 3y + 4 = 0 \) and a point
(i) \( (-2, 4) \)
(ii) \( (7, -3) \)
Answer: The distance \( d \) between a line \( ax + by + c = 0 \) and a point \( (x_1, y_1) \) is given by the formula:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
Here, the line is \( 4x + 3y + 4 = 0 \), so \( a = 4, b = 3, c = 4 \).

(i) For the point \( (x_1, y_1) = (-2, 4) \):
We plug the values into the distance formula:
\[ d = \frac{|4(-2) + 3(4) + 4|}{\sqrt{4^2 + 3^2}} \]
\[ d = \frac{|-8 + 12 + 4|}{\sqrt{16 + 9}} \]
\[ d = \frac{|8|}{\sqrt{25}} \]
\[ d = \frac{8}{5} \]

(ii) For the point \( (x_1, y_1) = (7, -3) \):
We use the same formula with the new point:
\[ d = \frac{|4(7) + 3(-3) + 4|}{\sqrt{4^2 + 3^2}} \]
\[ d = \frac{|28 - 9 + 4|}{\sqrt{16 + 9}} \]
\[ d = \frac{|23|}{\sqrt{25}} \]
\[ d = \frac{23}{5} \]
In simple words: To find the shortest distance from a point to a line, we use a special formula. We put the numbers from the line and the point into this formula and then calculate the result. This formula always gives the perpendicular distance.

🎯 Exam Tip: Always remember the absolute value sign in the numerator of the distance formula, as distance is always a positive quantity. Also, be careful with signs when substituting coordinates.

 

Question 4. Write the equation of the lines through the point \( (1, -1) \)
(i) Parallel to \( x + 3y - 4 = 0 \)
(ii) Perpendicular to \( 3x + 4y = 6 \)
Answer:
(i) For a line parallel to \( x + 3y - 4 = 0 \):
A line parallel to \( x + 3y - 4 = 0 \) will have the same slope, so its general equation is of the form \( x + 3y + k = 0 \).
We are given that this line passes through the point \( (1, -1) \). We can substitute these coordinates into the equation to find \( k \):
\( 1 + 3(-1) + k = 0 \)
\( 1 - 3 + k = 0 \)
\( -2 + k = 0 \)
\( k = 2 \)
Therefore, the equation of the required line is \( x + 3y + 2 = 0 \).

(ii) For a line perpendicular to \( 3x + 4y = 6 \):
The given line is \( 3x + 4y - 6 = 0 \). Its slope is \( m = - \frac{3}{4} \).
A line perpendicular to this will have a slope of \( - \frac{1}{m} \), which is \( - \frac{1}{(-3/4)} = \frac{4}{3} \).
The general equation of a line with slope \( \frac{4}{3} \) can be written as \( y = \frac{4}{3}x + c \), or by rearranging into the standard form for perpendicular lines: \( 4x - 3y + k = 0 \). This is a helpful shortcut: swap A and B and change the sign of one.
We are given that this line passes through the point \( (1, -1) \). Substitute these coordinates to find \( k \):
\( 4(1) - 3(-1) + k = 0 \)
\( 4 + 3 + k = 0 \)
\( 7 + k = 0 \)
\( k = -7 \)
Therefore, the equation of the required line is \( 4x - 3y - 7 = 0 \).
In simple words: For parallel lines, we use the same main part of the equation and find the last number using the given point. For perpendicular lines, we flip the numbers next to x and y, change a sign, and then find the last number using the given point.

🎯 Exam Tip: To find the equation of a line parallel to \( Ax + By + C = 0 \), use \( Ax + By + k = 0 \). For a line perpendicular to it, use \( Bx - Ay + k = 0 \) (or \( -Bx + Ay + k = 0 \)). Always substitute the given point to solve for \( k \).

 

Question 5. If \( (-4, 7) \) is one vertex of a rhombus and if the equation of one diagonal is \( 5x - y + 7 = 0 \), then find the equation of another diagonal.
Answer:
In a rhombus, the diagonals are perpendicular bisectors of each other. This means they cross at a right angle.
The equation of one diagonal is \( 5x - y + 7 = 0 \).
We check if the given vertex \( (-4, 7) \) lies on this diagonal by substituting its coordinates into the equation:
\( 5(-4) - (7) + 7 = -20 - 7 + 7 = -20 \).
Since \( -20 \neq 0 \), the point \( (-4, 7) \) does not lie on this diagonal. This means \( (-4, 7) \) must be a vertex on the *other* diagonal.
Since the diagonals of a rhombus are perpendicular, the other diagonal will be perpendicular to \( 5x - y + 7 = 0 \).
The equation of a line perpendicular to \( 5x - y + 7 = 0 \) is of the form \( x + 5y + k = 0 \). (We swap coefficients of x and y and change the sign of one).
This other diagonal passes through the vertex \( (-4, 7) \). Substitute these coordinates into the new equation to find \( k \):
\( (-4) + 5(7) + k = 0 \)
\( -4 + 35 + k = 0 \)
\( 31 + k = 0 \)
\( k = -31 \)
Therefore, the equation of the other diagonal is \( x + 5y - 31 = 0 \).
In simple words: Since the diagonals of a rhombus always cross each other at a right angle, we first find the equation of a line that is perpendicular to the given diagonal. Then, we use the fact that the vertex \( (-4, 7) \) must be on this new line to find its exact equation.

🎯 Exam Tip: Remember the key properties of a rhombus: diagonals are perpendicular bisectors. Use the point-on-line test to determine if a given vertex lies on a specific diagonal. The form \( Bx - Ay + k = 0 \) is handy for perpendicular lines to \( Ax + By + C = 0 \).

 

Question 6. Find the equation of the lines passing through the point of intersection of the lines \( 4x - y + 3 = 0 \) and \( 5x + 2y + 7 = 0 \) and
(i) through the point \( (-1, 2) \)
(ii) parallel to \( x - y + 5 = 0 \)
(iii) perpendicular to \( x - 2y + 1 = 0 \)
Answer: The general equation of a line passing through the intersection of two lines \( L_1 = 0 \) and \( L_2 = 0 \) is given by \( L_1 + \lambda L_2 = 0 \).
Here, \( L_1 = 4x - y + 3 \) and \( L_2 = 5x + 2y + 7 \). So the family of lines is:
\( (4x - y + 3) + \lambda (5x + 2y + 7) = 0 \quad \text{(1)} \)

(i) If the line passes through the point \( (-1, 2) \):
Substitute \( x = -1 \) and \( y = 2 \) into equation (1):
\( (4(-1) - 2 + 3) + \lambda (5(-1) + 2(2) + 7) = 0 \)
\( (-4 - 2 + 3) + \lambda (-5 + 4 + 7) = 0 \)
\( (-3) + \lambda (6) = 0 \)
\( 6\lambda = 3 \)
\( \lambda = \frac{3}{6} = \frac{1}{2} \)
Now, substitute \( \lambda = \frac{1}{2} \) back into equation (1):
\( (4x - y + 3) + \frac{1}{2} (5x + 2y + 7) = 0 \)
Multiply the entire equation by 2 to clear the fraction:
\( 2(4x - y + 3) + (5x + 2y + 7) = 0 \)
\( 8x - 2y + 6 + 5x + 2y + 7 = 0 \)
\( 13x + 13 = 0 \)
Divide by 13:
\( x + 1 = 0 \)

(ii) If the line is parallel to \( x - y + 5 = 0 \):
First, rearrange equation (1) to group x and y terms:
\( 4x - y + 3 + 5\lambda x + 2\lambda y + 7\lambda = 0 \)
\( (4 + 5\lambda)x + (-1 + 2\lambda)y + (3 + 7\lambda) = 0 \quad \text{(2)} \)
The slope of this line (2) is \( m_2 = - \frac{4 + 5\lambda}{-1 + 2\lambda} = \frac{4 + 5\lambda}{1 - 2\lambda} \).
The given line \( x - y + 5 = 0 \) has a slope \( m = - \frac{1}{-1} = 1 \).
For the lines to be parallel, their slopes must be equal:
\( \frac{4 + 5\lambda}{1 - 2\lambda} = 1 \)
\( 4 + 5\lambda = 1 - 2\lambda \)
\( 7\lambda = -3 \)
\( \lambda = - \frac{3}{7} \)
Substitute \( \lambda = - \frac{3}{7} \) back into equation (1):
\( (4x - y + 3) - \frac{3}{7} (5x + 2y + 7) = 0 \)
Multiply by 7:
\( 7(4x - y + 3) - 3(5x + 2y + 7) = 0 \)
\( 28x - 7y + 21 - 15x - 6y - 21 = 0 \)
\( 13x - 13y = 0 \)
Divide by 13:
\( x - y = 0 \)

(iii) If the line is perpendicular to \( x - 2y + 1 = 0 \):
The given line \( x - 2y + 1 = 0 \) has a slope \( m = - \frac{1}{-2} = \frac{1}{2} \).
For the lines to be perpendicular, the product of their slopes must be -1. So, the required slope for line (2) must be \( - \frac{1}{(1/2)} = -2 \).
Using the slope of line (2) found in part (ii):
\( \frac{4 + 5\lambda}{1 - 2\lambda} = -2 \)
\( 4 + 5\lambda = -2(1 - 2\lambda) \)
\( 4 + 5\lambda = -2 + 4\lambda \)
\( \lambda = -6 \)
Substitute \( \lambda = -6 \) back into equation (1):
\( (4x - y + 3) - 6(5x + 2y + 7) = 0 \)
\( 4x - y + 3 - 30x - 12y - 42 = 0 \)
\( -26x - 13y - 39 = 0 \)
Divide by -13:
\( 2x + y + 3 = 0 \)
In simple words: We start by writing a general equation for any line that passes through the meeting point of the two given lines. Then, we use the specific conditions (passing through another point, being parallel, or being perpendicular) to find the special value for \( \lambda \), which helps us get the final equation for each required line.

🎯 Exam Tip: The equation \( L_1 + \lambda L_2 = 0 \) is a powerful tool for finding lines passing through the intersection of two given lines. Remember that parallel lines have equal slopes (\( m_1 = m_2 \)), and perpendicular lines have slopes whose product is -1 (\( m_1 m_2 = -1 \)).

 

Question 7. Find the equations of two straight lines which are parallel to the line \( 12x + 5y + 2 = 0 \) and at a unit distance from the point \( (1, -1) \).
Answer: The given line is \( 12x + 5y + 2 = 0 \).
Any line parallel to this line will have the same slope, so its equation can be written in the form:
\( 12x + 5y + k = 0 \quad \text{(1)} \)
We are given that this line is at a unit distance (distance = 1) from the point \( (1, -1) \).
The distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is given by the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 12, B = 5, C = k \), and \( (x_1, y_1) = (1, -1) \). We set \( d = 1 \):
\[ 1 = \frac{|12(1) + 5(-1) + k|}{\sqrt{12^2 + 5^2}} \]
\[ 1 = \frac{|12 - 5 + k|}{\sqrt{144 + 25}} \]
\[ 1 = \frac{|7 + k|}{\sqrt{169}} \]
\[ 1 = \frac{|7 + k|}{13} \]
Now, we solve for \( k \):
\( |7 + k| = 13 \)
This gives two possibilities:
\( 7 + k = 13 \quad \text{or} \quad 7 + k = -13 \)
For the first case:
\( k = 13 - 7 \Rightarrow k = 6 \)
For the second case:
\( k = -13 - 7 \Rightarrow k = -20 \)
So there are two possible values for \( k \), which means there are two such lines.
Substitute these values of \( k \) back into the general equation \( 12x + 5y + k = 0 \):
1. When \( k = 6 \): \( 12x + 5y + 6 = 0 \)
2. When \( k = -20 \): \( 12x + 5y - 20 = 0 \)
These are the two required equations of straight lines. These lines are like two train tracks, both parallel to the original track, one on each side at the specified distance.
In simple words: We first set up a general equation for lines that are parallel to the given line. Then, we used the distance formula to figure out the exact numbers (k values) that would make these lines exactly one unit away from the given point. This gave us two possible lines.

🎯 Exam Tip: When dealing with absolute values like \( |X| = a \), remember that \( X = a \) or \( X = -a \), leading to two possible solutions. This often means there are two lines that satisfy the given distance condition.

 

Question 8. Find the equations of straight lines which are perpendicular to the line \( 3x + 4y - 6 = 0 \) and are at a distance of 4 units from \( (2, 1) \).
Answer: The given line is \( 3x + 4y - 6 = 0 \).
A line perpendicular to \( Ax + By + C = 0 \) is of the form \( Bx - Ay + k = 0 \).
So, a line perpendicular to \( 3x + 4y - 6 = 0 \) will have the equation:
\( 4x - 3y + k = 0 \quad \text{(1)} \)
We are given that this line is at a distance of 4 units from the point \( (2, 1) \).
Using the distance formula \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \):
Here, \( A = 4, B = -3, C = k \), and \( (x_1, y_1) = (2, 1) \). We set \( d = 4 \):
\[ 4 = \frac{|4(2) - 3(1) + k|}{\sqrt{4^2 + (-3)^2}} \]
\[ 4 = \frac{|8 - 3 + k|}{\sqrt{16 + 9}} \]
\[ 4 = \frac{|5 + k|}{\sqrt{25}} \]
\[ 4 = \frac{|5 + k|}{5} \]
Now, solve for \( k \):
\( |5 + k| = 4 \times 5 \)
\( |5 + k| = 20 \)
This gives two possibilities:
\( 5 + k = 20 \quad \text{or} \quad 5 + k = -20 \)
For the first case:
\( k = 20 - 5 \Rightarrow k = 15 \)
For the second case:
\( k = -20 - 5 \Rightarrow k = -25 \)
Substitute these values of \( k \) back into the general equation \( 4x - 3y + k = 0 \):
1. When \( k = 15 \): \( 4x - 3y + 15 = 0 \)
2. When \( k = -25 \): \( 4x - 3y - 25 = 0 \)
These are the two required equations of straight lines. There are always two such lines, one on each side of the point, at the specified perpendicular distance.
In simple words: First, we wrote down the general look of a line that is perpendicular to the given line. Then, we used the distance formula to find the exact two lines that are exactly 4 units away from the given point.

🎯 Exam Tip: Remember to consider both positive and negative cases when solving for \( k \) after taking the absolute value, as distances are always positive but the expression inside the absolute value can be positive or negative.

 

Question 9. Find the equation of a straight line parallel to \( 2x + 3y = 10 \) and which is such that the sum of its intercepts on the axes is 15.
Answer: The given line is \( 2x + 3y = 10 \).
A line parallel to \( 2x + 3y = 10 \) will have the same slope, so its equation can be written as:
\( 2x + 3y = k \quad \text{(1)} \)
To find the intercepts on the axes, we can convert this equation to the intercept form, which is \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) is the x-intercept and \( b \) is the y-intercept.
Divide equation (1) by \( k \):
\( \frac{2x}{k} + \frac{3y}{k} = 1 \)
\( \frac{x}{k/2} + \frac{y}{k/3} = 1 \)
From this, the x-intercept is \( a = \frac{k}{2} \) and the y-intercept is \( b = \frac{k}{3} \).
We are given that the sum of its intercepts on the axes is 15:
\( a + b = 15 \)
\( \frac{k}{2} + \frac{k}{3} = 15 \)
To solve for \( k \), find a common denominator for the fractions (which is 6):
\( \frac{3k + 2k}{6} = 15 \)
\( \frac{5k}{6} = 15 \)
\( 5k = 15 \times 6 \)
\( 5k = 90 \)
\( k = \frac{90}{5} \)
\( k = 18 \)
Substitute the value of \( k \) back into the general equation (1):
\( 2x + 3y = 18 \)
This is the required equation of the straight line. This line has its x and y intercepts summing up to 15, while maintaining its parallel orientation to the original line.
In simple words: We first set up a general equation for a line parallel to the given one. Then, we found where this general line crosses the x and y axes. Since we knew these crossing points (intercepts) add up to 15, we could find the missing number in our line's equation.

🎯 Exam Tip: When dealing with intercepts, convert the line equation into the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \). This directly gives the x-intercept \( a \) and y-intercept \( b \), making calculations involving their sum or product straightforward.

 

Question 10. Find the length of the perpendicular and the coordinates of the foot of the perpendicular from \( (-10, -2) \) to the line \( x + y - 2 = 0 \).
Answer: Let the given point be \( P(x_1, y_1) = (-10, -2) \) and the line be \( x + y - 2 = 0 \), so \( a = 1, b = 1, c = -2 \).

**1. Find the coordinates of the foot of the perpendicular:**
The formula for the coordinates of the foot of the perpendicular \( (x_f, y_f) \) from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is:
\[ \frac{x_f - x_1}{a} = \frac{y_f - y_1}{b} = - \frac{ax_1 + by_1 + c}{a^2 + b^2} \]
Substitute the given values:
\[ \frac{x_f - (-10)}{1} = \frac{y_f - (-2)}{1} = - \frac{1(-10) + 1(-2) - 2}{1^2 + 1^2} \]
\[ \frac{x_f + 10}{1} = \frac{y_f + 2}{1} = - \frac{-10 - 2 - 2}{1 + 1} \]
\[ \frac{x_f + 10}{1} = \frac{y_f + 2}{1} = - \frac{-14}{2} \]
\[ \frac{x_f + 10}{1} = \frac{y_f + 2}{1} = 7 \]
Now, we equate each part to 7:
\( x_f + 10 = 7 \implies x_f = 7 - 10 \implies x_f = -3 \)
\( y_f + 2 = 7 \implies y_f = 7 - 2 \implies y_f = 5 \)
So, the coordinates of the foot of the perpendicular are \( (-3, 5) \). This point is the closest point on the line to the given point.

**2. Find the length of the perpendicular:**
The length of the perpendicular is the distance between the original point \( P(-10, -2) \) and the foot of the perpendicular \( F(-3, 5) \). We use the distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \):
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
\[ d = \sqrt{(-3 - (-10))^2 + (5 - (-2))^2} \]
\[ d = \sqrt{(-3 + 10)^2 + (5 + 2)^2} \]
\[ d = \sqrt{(7)^2 + (7)^2} \]
\[ d = \sqrt{49 + 49} \]
\[ d = \sqrt{98} \]
We can simplify \( \sqrt{98} \): \( \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2} \)
So, the length of the perpendicular is \( 7\sqrt{2} \) units.
Alternatively, we could have used the point-to-line distance formula directly for the length: \( d = \frac{|1(-10) + 1(-2) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|-10 - 2 - 2|}{\sqrt{1+1}} = \frac{|-14|}{\sqrt{2}} = \frac{14}{\sqrt{2}} = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \).
In simple words: First, we found the exact spot on the line where a straight path from the point would hit it at a perfect right angle. This spot is called the foot of the perpendicular. Then, we simply measured the distance between our starting point and this special spot on the line.

🎯 Exam Tip: You can calculate the length of the perpendicular using the point-to-line distance formula directly, which is often quicker than finding the foot of the perpendicular first and then using the distance between two points. However, if asked for both, calculate the foot first.

 

Question 11. If \( p_1 \) and \( p_2 \) are the lengths of the perpendiculars from the origin to the straight lines \( x \sec \theta + y \csc \theta = 2a \) and \( x \cos \theta - y \sin \theta = a \cos 2\theta \), then prove that \( p_1^2 + p_2^2 = a^2 \).
Answer: The distance \( d \) from the origin \( (0,0) \) to a line \( Ax + By + C = 0 \) is given by \( d = \frac{|C|}{\sqrt{A^2 + B^2}} \).

**For \( p_1 \):**
The first line is \( x \sec \theta + y \csc \theta - 2a = 0 \).
Here, \( A = \sec \theta, B = \csc \theta, C = -2a \).
So, the length of the perpendicular \( p_1 \) from the origin is:
\[ p_1 = \frac{|-2a|}{\sqrt{\sec^2 \theta + \csc^2 \theta}} = \frac{2|a|}{\sqrt{\sec^2 \theta + \csc^2 \theta}} \]
Square both sides to find \( p_1^2 \):
\[ p_1^2 = \frac{4a^2}{\sec^2 \theta + \csc^2 \theta} \]
We know that \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \) and \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \). Substitute these identities:
\[ p_1^2 = \frac{4a^2}{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}} \]
Combine the terms in the denominator:
\[ p_1^2 = \frac{4a^2}{\frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta}} \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ p_1^2 = \frac{4a^2}{\frac{1}{\cos^2 \theta \sin^2 \theta}} \]
\[ p_1^2 = 4a^2 \cos^2 \theta \sin^2 \theta \]
We know the double angle identity \( \sin 2\theta = 2 \sin \theta \cos \theta \). So, \( (2 \sin \theta \cos \theta)^2 = \sin^2 2\theta \).
\[ p_1^2 = a^2 (2 \sin \theta \cos \theta)^2 = a^2 \sin^2 2\theta \quad \text{(Equation A)} \]

**For \( p_2 \):**
The second line is \( x \cos \theta - y \sin \theta - a \cos 2\theta = 0 \).
Here, \( A = \cos \theta, B = -\sin \theta, C = -a \cos 2\theta \).
So, the length of the perpendicular \( p_2 \) from the origin is:
\[ p_2 = \frac{|-a \cos 2\theta|}{\sqrt{(\cos \theta)^2 + (-\sin \theta)^2}} \]
\[ p_2 = \frac{|-a \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ p_2 = \frac{|-a \cos 2\theta|}{\sqrt{1}} = |-a \cos 2\theta| \]
Square both sides to find \( p_2^2 \):
\[ p_2^2 = (-a \cos 2\theta)^2 = a^2 \cos^2 2\theta \quad \text{(Equation B)} \]

**Now, we need to prove \( p_1^2 + p_2^2 = a^2 \):**
Add Equation A and Equation B:
\( p_1^2 + p_2^2 = a^2 \sin^2 2\theta + a^2 \cos^2 2\theta \)
Factor out \( a^2 \):
\( p_1^2 + p_2^2 = a^2 (\sin^2 2\theta + \cos^2 2\theta) \)
Using the trigonometric identity \( \sin^2 X + \cos^2 X = 1 \), where \( X = 2\theta \):
\( p_1^2 + p_2^2 = a^2 (1) \)
\( p_1^2 + p_2^2 = a^2 \)
Hence, it is proven. This shows a beautiful relationship between perpendicular distances from the origin and trigonometric identities.
In simple words: We used a formula to find the squared distance of two lines from the center point (origin). After using some basic math rules about angles (trigonometric identities), we added these two squared distances together and showed that the answer is always \( a^2 \), just as the question asked.

🎯 Exam Tip: This problem is a common application of the distance formula from a point to a line and fundamental trigonometric identities. Remember that \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( \sin 2\theta = 2 \sin \theta \cos \theta \), as these are crucial for simplifying the expressions.

 

Question 12. Find the distance between the parallel lines
(i) \( 12x + 5y = 7 \) and \( 12x + 5y + 7 = 0 \)
(ii) \( 3x - 4y + 5 = 0 \) and \( 6x - 8y - 15 = 0 \)
Answer: The distance \( d \) between two parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is given by the formula:
\[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \]

(i) For the lines \( 12x + 5y = 7 \) and \( 12x + 5y + 7 = 0 \):
First, rewrite the first equation in the standard form \( Ax + By + C = 0 \):
\( 12x + 5y - 7 = 0 \)
Now, we have \( A = 12, B = 5 \). The constants are \( C_1 = -7 \) and \( C_2 = 7 \).
Substitute these values into the distance formula:
\[ d = \frac{|-7 - 7|}{\sqrt{12^2 + 5^2}} \]
\[ d = \frac{|-14|}{\sqrt{144 + 25}} \]
\[ d = \frac{14}{\sqrt{169}} \]
\[ d = \frac{14}{13} \]

(ii) For the lines \( 3x - 4y + 5 = 0 \) and \( 6x - 8y - 15 = 0 \):
For the distance formula to apply, the coefficients of \( x \) and \( y \) (A and B) must be the same for both equations. We can divide the second equation by 2:
\( \frac{6x}{2} - \frac{8y}{2} - \frac{15}{2} = 0 \)
\( 3x - 4y - \frac{15}{2} = 0 \)
Now, both lines have \( A = 3, B = -4 \). The constants are \( C_1 = 5 \) and \( C_2 = - \frac{15}{2} \).
Substitute these values into the distance formula:
\[ d = \frac{|5 - (-\frac{15}{2})|}{\sqrt{3^2 + (-4)^2}} \]
\[ d = \frac{|5 + \frac{15}{2}|}{\sqrt{9 + 16}} \]
\[ d = \frac{|\frac{10}{2} + \frac{15}{2}|}{\sqrt{25}} \]
\[ d = \frac{|\frac{25}{2}|}{5} \]
\[ d = \frac{25/2}{5} \]
\[ d = \frac{25}{2 \times 5} \]
\[ d = \frac{5}{2} \]
In simple words: To find the distance between two parallel lines, we first make sure the numbers in front of 'x' and 'y' are the same for both lines. Then, we use a special formula that involves the constant numbers from each line and the square root of the squared numbers in front of 'x' and 'y'.

🎯 Exam Tip: Before applying the distance formula for parallel lines, always ensure that the coefficients of \( x \) and \( y \) are identical in both equations. If they are not, multiply or divide one of the equations by a suitable constant to make them match.

 

Question 13. Find the family of straight lines:
(i) Perpendicular to \( 3x + 4y - 12 = 0 \)
(ii) Parallel to \( 3x + 4y - 12 = 0 \)
Answer:
(i) The family of lines perpendicular to \( 3x + 4y - 12 = 0 \):
For a line \( Ax + By + C = 0 \), a perpendicular line can be written in the form \( Bx - Ay + k = 0 \).
So, for the line \( 3x + 4y - 12 = 0 \), a perpendicular line will be of the form \( 4x - 3y + k = 0 \).
Here, \( k \) is an arbitrary constant, meaning any real number. This constant determines the specific position of the line in the family.
Therefore, the family of lines perpendicular to \( 3x + 4y - 12 = 0 \) is \( 4x - 3y + k = 0 \), where \( k \in \mathbb{R} \).

(ii) The family of lines parallel to \( 3x + 4y - 12 = 0 \):
For a line \( Ax + By + C = 0 \), a parallel line can be written in the form \( Ax + By + k = 0 \).
So, for the line \( 3x + 4y - 12 = 0 \), a parallel line will be of the form \( 3x + 4y + k = 0 \).
Again, \( k \) is an arbitrary constant that determines the specific line within the family.
Therefore, the family of lines parallel to \( 3x + 4y - 12 = 0 \) is \( 3x + 4y + k = 0 \), where \( k \in \mathbb{R} \).
In simple words: We find the general form for lines that are either perfectly straight alongside a given line (parallel) or perfectly crossed at a right angle (perpendicular). The letter 'k' stands for any number, showing that there are many such lines in each family.

🎯 Exam Tip: Understanding the general forms for parallel (\( Ax + By + k = 0 \)) and perpendicular (\( Bx - Ay + k = 0 \)) lines is crucial. The constant \( k \) represents the entire family of such lines, each with a different intercept or position.

 

Question 14. If the line joining two points \( A(2, 0) \) and \( B(3, 1) \) is rotated about A in an anticlockwise direction through an angle of \( 15^\circ \), then find the equation of the line in the new position.
Answer:
First, we find the slope of the original line AB joining \( A(2,0) \) and \( B(3,1) \).
The slope \( m \) is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \):
\( m_{AB} = \frac{1 - 0}{3 - 2} = \frac{1}{1} = 1 \)
The angle \( \theta \) that the line AB makes with the positive x-axis is found using \( m = \tan \theta \):
\( \tan \theta = 1 \implies \theta = 45^\circ \)
The line AB is rotated about point A in an anticlockwise direction through an angle of \( 15^\circ \).
The new angle \( \theta' \) that the rotated line AB' makes with the x-axis will be the original angle plus the rotation angle:
\( \theta' = 45^\circ + 15^\circ = 60^\circ \)
The slope of the new line AB' will be \( m' = \tan \theta' \):
\( m' = \tan 60^\circ = \sqrt{3} \)
Now we have the slope \( m' = \sqrt{3} \) and a point \( A(2,0) \) through which the new line passes. We use the point-slope form of a line: \( y - y_1 = m(x - x_1) \).
\( y - 0 = \sqrt{3} (x - 2) \)
\( y = \sqrt{3}x - 2\sqrt{3} \)
Rearrange into the standard form:
\( \sqrt{3}x - y - 2\sqrt{3} = 0 \)
This is the equation of the line in its new position. Rotation changes the orientation of the line while keeping one point fixed.
In simple words: We first found the starting angle of the line with the x-axis. Since the line was spun around a point by 15 degrees, we added 15 degrees to its original angle to find its new angle. Then, using this new angle and the point it spun around, we wrote the equation for the line in its new spot.

🎯 Exam Tip: When a line is rotated about a point, its slope changes while the rotation point remains on the line. The new angle is found by adding/subtracting the rotation angle from the original angle, and the point-slope form \( y - y_1 = m(x - x_1) \) is ideal for finding the equation.

 

Question 15. A ray of light coming from the point \( (1, 2) \) is reflected at a point A on the x-axis and it passes through the point \( (5, 3) \). Find the coordinates of point A.
Answer:
Let the point on the x-axis where the reflection occurs be \( A(x, 0) \).
Let the incident ray come from \( P(1, 2) \) and the reflected ray pass through \( B(5, 3) \).
According to the law of reflection, the angle of incidence equals the angle of reflection. When reflection occurs on the x-axis, the angle the incident ray makes with the x-axis (say, \( \theta_1 \)) and the angle the reflected ray makes with the x-axis (say, \( \theta_2 \)) are related by \( \tan \theta_1 = - \tan \theta_2 \). This means if the incident ray has slope \( m_1 \), the reflected ray will have slope \( m_2 \), such that \( m_1 = -m_2 \) if measured from the point of reflection.

Alternatively, consider the image of point \( P(1, 2) \) across the x-axis. Let this image be \( P'(1, -2) \). The reflected ray from P to A and then to B will appear to come from \( P' \) in a straight line to A and then to B. Thus, points \( P' \), \( A \), and \( B \) are collinear.
For three points to be collinear, the slope between any two pairs of points must be the same.

Slope of \( P'A = \frac{0 - (-2)}{x - 1} = \frac{2}{x - 1} \)
Slope of \( AB = \frac{3 - 0}{5 - x} = \frac{3}{5 - x} \)

Since \( P', A, B \) are collinear, their slopes must be equal:
\( \frac{2}{x - 1} = \frac{3}{5 - x} \)
Now, cross-multiply and solve for \( x \):
\( 2(5 - x) = 3(x - 1) \)
\( 10 - 2x = 3x - 3 \)
\( 10 + 3 = 3x + 2x \)
\( 13 = 5x \)
\( x = \frac{13}{5} \)
So, the coordinates of point A on the x-axis are \( (\frac{13}{5}, 0) \). This is the point on the x-axis where the light ray is reflected.
In simple words: When light bounces off a flat surface (like the x-axis), it acts as if it came from a "mirror image" of its starting point. So, we found the mirror image of the first point, then drew a straight line from this mirror image through the reflection point to the final point. By making these three points line up, we found the exact spot on the x-axis where the light bounced.

🎯 Exam Tip: For reflection problems across an axis, a powerful technique is to use the concept of an image point. Reflect the incident point across the axis of reflection; the image point, the point of reflection on the axis, and the reflected point will be collinear. Then, use the collinearity condition (equal slopes) to find the unknown coordinate.

 

Question 16. A line is drawn perpendicular to 5x = y + 7. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 10 sq. units.
Answer: First, we write the given line's equation as \( 5x - y - 7 = 0 \). Let's call this Line (1). A line perpendicular to Line (1) will have the form \( x + 5y - k = 0 \). Let's call this Line AB. This line forms a triangle with the coordinate axes. The x-intercept is found by setting \( y = 0 \), which gives \( x = k \). So point A is \( (k, 0) \). The y-intercept is found by setting \( x = 0 \), which gives \( 5y = k \), so \( y = \frac{k}{5} \). So point B is \( (0, \frac{k}{5}) \). The origin is O \( (0, 0) \).
The area of triangle OAB is given by the formula \( \frac{1}{2} \times \text{base} \times \text{height} \). Here, the base is \( OA = |k| \) and the height is \( OB = |\frac{k}{5}| \).
So, Area \( = \frac{1}{2} \times |k| \times |\frac{k}{5}| = \frac{k^2}{10} \).
We are given that the area is 10 sq. units.
\( \frac{k^2}{10} = 10 \)
\( \implies k^2 = 100 \)
\( \implies k = \pm 10 \)
Therefore, the required equation of the straight line is \( x + 5y = \pm 10 \). This means there are two possible lines: \( x + 5y - 10 = 0 \) and \( x + 5y + 10 = 0 \).

In simple words: We find the type of line that is at right angles to the first line. Then we see where this new line cuts the 'x' and 'y' lines on a graph. These cutting points, with the middle point (origin), form a triangle. We use the rule for the area of this triangle to find the missing number 'k' in our line's equation. Since area is positive, 'k' can be both positive or negative, giving two possible lines.

🎯 Exam Tip: Remember that when finding the equation of a line perpendicular to \( Ax + By + C = 0 \), the new equation will be of the form \( Bx - Ay + k = 0 \) (or \( -Bx + Ay + k = 0 \)). Also, always consider both positive and negative values when taking the square root in geometry problems, as distances can be represented by either, but squared terms resolve to positive.

 

Question 17. Find the image of the point (- 2, 3) about the line x + 2y – 9 = 0.
Answer: We need to find the reflection of a point \( (x_1, y_1) \) across a line \( ax + by + c = 0 \). The formula for the image \( (x, y) \) is:
\( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2} \)
Here, the given point is \( (x_1, y_1) = (-2, 3) \).
The given line is \( x + 2y - 9 = 0 \), so \( a = 1 \), \( b = 2 \), \( c = -9 \).
Now, let's plug these values into the formula:
\( \frac{x - (-2)}{1} = \frac{y - 3}{2} = \frac{-2(1(-2) + 2(3) - 9)}{1^2 + 2^2} \)
First, calculate the right side of the equation:
\( \frac{-2(-2 + 6 - 9)}{1 + 4} = \frac{-2(-5)}{5} = \frac{10}{5} = 2 \)
So, we have:
\( \frac{x + 2}{1} = 2 \)
\( \implies x + 2 = 2 \)
\( \implies x = 0 \)
And:
\( \frac{y - 3}{2} = 2 \)
\( \implies y - 3 = 4 \)
\( \implies y = 7 \)
Thus, the image of the point \( (-2, 3) \) about the line \( x + 2y - 9 = 0 \) is \( (0, 7) \).
In simple words: Imagine the line as a mirror. We want to find where the point would appear in that mirror. We use a special formula that calculates this mirror image point based on the original point's location and the line's equation. After putting in all the numbers, we find the new coordinates for the reflected point.

🎯 Exam Tip: Carefully substitute the coordinates of the point and the coefficients of the line into the formula. A common mistake is to forget the \( -2 \) factor in the numerator of the right-hand side, or to miscalculate \( a^2 + b^2 \).

 

Question 18. A photocopy store charges Rs. 1.50 per copy for the first 10 copies and Rs. 1.00 per copy after the 10th copy. Let x be the number of copies, and y be the total cost of photo coping.
(i) Draw graph the cost as x goes from 0 to 50 copies
(ii) Find the cost of making 40 copies.
Answer:
(i) Let's find the cost function \( y \) based on the number of copies \( x \).
For the first 10 copies (when \( 0 \le x \le 10 \)):
Cost per copy = Rs. 1.50
So, \( y = 1.50x \)
For copies after the 10th (when \( x > 10 \)):
Cost for the first 10 copies = \( 10 \times 1.50 = 15 \) Rs.
Cost for the remaining \( (x - 10) \) copies = \( (x - 10) \times 1.00 = x - 10 \) Rs.
Total cost \( y = 15 + (x - 10) \)
So, \( y = x + 5 \)
Combining these, the cost function is:
\[ y = \begin{cases} 1.50x & \text{if } 0 \le x \le 10 \\ x + 5 & \text{if } x > 10 \end{cases} \]
To draw the graph from \( x = 0 \) to \( x = 50 \), we can find points for plotting:
When \( x = 0 \), \( y = 1.50 \times 0 = 0 \). Point: \( (0, 0) \).
When \( x = 10 \), \( y = 1.50 \times 10 = 15 \). Point: \( (10, 15) \).
For \( x > 10 \), use \( y = x + 5 \):
When \( x = 20 \), \( y = 20 + 5 = 25 \). Point: \( (20, 25) \).
When \( x = 30 \), \( y = 30 + 5 = 35 \). Point: \( (30, 35) \).
When \( x = 40 \), \( y = 40 + 5 = 45 \). Point: \( (40, 45) \).
When \( x = 50 \), \( y = 50 + 5 = 55 \). Point: \( (50, 55) \).
The graph will be a line segment from \( (0, 0) \) to \( (10, 15) \), and then another line segment from \( (10, 15) \) to \( (50, 55) \). The first segment is steeper than the second, as its slope is 1.5 compared to 1. This type of graph is common for pricing that changes after a certain quantity.

(ii) To find the cost of making 40 copies, we use the cost function for \( x > 10 \), which is \( y = x + 5 \).
Substitute \( x = 40 \):
\( y = 40 + 5 \)
\( y = 45 \)
So, the cost of making 40 copies is Rs. 45.
In simple words: For the first part, we figured out how the cost changes based on how many copies you make. It's cheaper per copy after the first ten. We made a list of costs for different numbers of copies to show how the price goes up. For the second part, we just used the rule for copies over ten to quickly find out the cost for 40 copies.

🎯 Exam Tip: When dealing with piecewise functions (functions with different rules for different ranges), always clearly define each part of the function. For graph drawing, calculate points for both sections and ensure the graph shows the change in slope at the transition point. For calculations, identify the correct function rule based on the given input value.

 

Question 19. Find at least two equations of the straight lines in the family of lines y = 5x + b for which b and the x - coordinate of the point of intersection of the lines with 3x – 4y = 6 are integers.
Answer: We are given two lines:
Line (1): \( y = 5x + b \)
Line (2): \( 3x - 4y = 6 \)
We need to find the point of intersection. Substitute \( y \) from Line (1) into Line (2):
\( 3x - 4(5x + b) = 6 \)
\( 3x - 20x - 4b = 6 \)
\( -17x = 6 + 4b \)
\( x = \frac{6 + 4b}{-17} \) or \( x = -\frac{6 + 4b}{17} \)
For \( x \) to be an integer, \( (6 + 4b) \) must be a multiple of 17. We need to find integer values of \( b \) that make \( (6 + 4b) \) a multiple of 17.
Let's test some integer values for \( b \):
Case 1: Let \( b = 7 \)
Then, \( 6 + 4b = 6 + 4(7) = 6 + 28 = 34 \).
Since 34 is a multiple of 17 ( \( 34 = 2 \times 17 \) ), this works.
\( x = \frac{34}{-17} = -2 \)
So, for \( b = 7 \), \( x = -2 \) (an integer).
The equation of the line is \( y = 5x + 7 \).
Case 2: Let \( b = -10 \)
Then, \( 6 + 4b = 6 + 4(-10) = 6 - 40 = -34 \).
Since -34 is a multiple of 17 ( \( -34 = -2 \times 17 \) ), this also works.
\( x = \frac{-34}{-17} = 2 \)
So, for \( b = -10 \), \( x = 2 \) (an integer).
The equation of the line is \( y = 5x - 10 \).
Thus, two required equations are \( y = 5x + 7 \) and \( y = 5x - 10 \). These lines belong to the family and have integer x-coordinates for their intersection points with the other given line.
In simple words: We have a group of lines that look like \( y = 5x + b \), where 'b' can be any number. We also have another fixed line. We want to find two lines from the first group such that when they cross the fixed line, the 'x' part of their crossing point is a whole number, and 'b' is also a whole number. We do this by solving the equations to find 'x' in terms of 'b'. Then we test different whole numbers for 'b' until 'x' also becomes a whole number. This gives us two such lines.

🎯 Exam Tip: When asked to find integer solutions, first express the variable (like \( x \) in this case) in terms of the parameter (like \( b \)). Then, use trial and error or modular arithmetic to find integer values for the parameter that make the expression an integer. Always verify your chosen values.

 

Question 20. Find all the equations of the straight lines in the family of the lines y = mx – 3 for which m and the x coordinate of the point of intersection of the lines with x – y = 6 are integers.
Answer: We are given two lines:
Line (1): \( y = mx - 3 \)
Line (2): \( x - y = 6 \)
We need to find the point of intersection. Substitute \( y \) from Line (1) into Line (2):
\( x - (mx - 3) = 6 \)
\( x - mx + 3 = 6 \)
\( x(1 - m) = 6 - 3 \)
\( x(1 - m) = 3 \)
\( x = \frac{3}{1 - m} \)
For \( x \) to be an integer, \( (1 - m) \) must be an integer divisor of 3. The integer divisors of 3 are \( \{ -3, -1, 1, 3 \} \). We also know that \( m \) must be an integer.
Let's consider each possible value for \( (1 - m) \):
Case 1: If \( 1 - m = 1 \)
\( m = 1 - 1 = 0 \)
Then, \( x = \frac{3}{1} = 3 \)
Since both \( m=0 \) and \( x=3 \) are integers, this is a valid solution.
The corresponding equation is \( y = (0)x - 3 \), which simplifies to \( y = -3 \), or \( y + 3 = 0 \).
Case 2: If \( 1 - m = -1 \)
\( m = 1 - (-1) = 1 + 1 = 2 \)
Then, \( x = \frac{3}{-1} = -3 \)
Since both \( m=2 \) and \( x=-3 \) are integers, this is a valid solution.
The corresponding equation is \( y = (2)x - 3 \), which is \( y = 2x - 3 \). This can also be written as \( 2x - y - 3 = 0 \).
Case 3: If \( 1 - m = 3 \)
\( m = 1 - 3 = -2 \)
Then, \( x = \frac{3}{3} = 1 \)
Since both \( m=-2 \) and \( x=1 \) are integers, this is a valid solution.
The corresponding equation is \( y = (-2)x - 3 \), which is \( y = -2x - 3 \). This can also be written as \( 2x + y + 3 = 0 \).
Therefore, the three equations of the straight lines that satisfy the given conditions are: \( y + 3 = 0 \), \( 2x - y - 3 = 0 \), and \( 2x + y + 3 = 0 \). Each of these lines belongs to the family \( y = mx - 3 \) and results in integer values for both \( m \) and the x-coordinate of the intersection point.
In simple words: We have a group of lines that look like \( y = mx - 3 \), where 'm' can be any number. We also have another fixed line. We want to find all lines from the first group such that when they cross the fixed line, both the 'm' value and the 'x' part of their crossing point are whole numbers. We do this by finding 'x' in terms of 'm' and then figuring out which whole numbers for 'm' make 'x' also a whole number. This gives us three such lines.

🎯 Exam Tip: When the problem states that both a parameter (like \( m \)) and a coordinate (like \( x \)) must be integers, remember to list all integer divisors of the numerator for the expression defining the coordinate. Each divisor will lead to a valid integer value for the parameter, providing all possible solutions.