OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Exercise 16 (H)

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(h)

Question 1. Find the equation of the straight line which passes through the point (4, 5) and is (i) parallel, (ii) perpendicular to the straight line \( 3x - 2y + 5 = 0 \).
Answer:
(i) To find the equation of a line parallel to \( 3x - 2y + 5 = 0 \) and passing through (4, 5):
The given equation of the straight line is \( 3x - 2y + 5 = 0 \) ...(1)
A line parallel to line (1) will have the form \( 3x - 2y + k = 0 \) ...(2)
Since equation (2) passes through the point (4, 5), we substitute x = 4 and y = 5 into it:
\( 3(4) - 2(5) + k = 0 \)
\( 12 - 10 + k = 0 \)
\( 2 + k = 0 \)
\( k = -2 \)
Now, we put the value of \( k = -2 \) back into equation (2):
\( 3x - 2y - 2 = 0 \)
This is the required equation of the line. Parallel lines have the same slope, so only the constant term changes.

(ii) To find the equation of a line perpendicular to \( 3x - 2y + 5 = 0 \) and passing through (4, 5):
The given equation of the straight line is \( 3x - 2y + 5 = 0 \) ...(1)
A line perpendicular to line (1) will have the form \( 2x + 3y + c = 0 \) ...(3) (Here, the coefficients of x and y are swapped, and the sign of one is changed).
Since equation (3) passes through the point (4, 5), we substitute x = 4 and y = 5 into it:
\( 2(4) + 3(5) + c = 0 \)
\( 8 + 15 + c = 0 \)
\( 23 + c = 0 \)
\( c = -23 \)
Now, we put the value of \( c = -23 \) back into equation (3):
\( 2x + 3y - 23 = 0 \)
This is the required equation of the line. The product of slopes of perpendicular lines is -1.
In simple words: First, we find the line that is going the same way as the given line and goes through (4, 5). Then, we find another line that crosses the given line at a perfect right angle (90 degrees) and also goes through (4, 5).

🎯 Exam Tip: Remember that if a line is \( Ax + By + C = 0 \), a parallel line is \( Ax + By + k = 0 \), and a perpendicular line is \( Bx - Ay + c = 0 \) (or \( -Bx + Ay + c = 0 \)). Always substitute the given point to find the constant \( k \) or \( c \).

Question 2. Find the equation of the straight line which is such that
(i) it passes through the point (4, 3) and is parallel to the line \( 3x - 4y + 5 = 0 \).
(ii) it passes through the point (4, 3) and is perpendicular to the line \( 3x - 4y + 5 = 0 \).
Answer:
(i) To find the equation of a line parallel to \( 3x - 4y + 5 = 0 \) and passing through (4, 3):
The given equation of the straight line is \( 3x - 4y + 5 = 0 \) ...(1)
The equation of a line parallel to line (1) will be \( 3x - 4y + k = 0 \) ...(2)
Since equation (2) passes through the point (4, 3), we substitute x = 4 and y = 3 into it:
\( 3(4) - 4(3) + k = 0 \)
\( 12 - 12 + k = 0 \)
\( k = 0 \)
Now, we put the value of \( k = 0 \) back into equation (2):
\( 3x - 4y + 0 = 0 \)
\( 3x - 4y = 0 \)
This is the required equation of the line. This means the line passes through the origin.

(ii) To find the equation of a line perpendicular to \( 3x - 4y + 5 = 0 \) and passing through (4, 3):
The given equation of the straight line is \( 3x - 4y + 5 = 0 \).
The equation of a line perpendicular to this line will be \( 4x + 3y + k = 0 \).
Since this line passes through the point (4, 3), we substitute x = 4 and y = 3 into it:
\( 4(4) + 3(3) + k = 0 \)
\( 16 + 9 + k = 0 \)
\( 25 + k = 0 \)
\( k = -25 \)
Now, we put the value of \( k = -25 \) back into the equation:
\( 4x + 3y - 25 = 0 \)
This is the required equation of the line. The slopes of perpendicular lines have a product of -1.
In simple words: We need to find two lines. The first one runs in the same direction as the given line and passes through the point (4, 3). The second one cuts the given line at a right angle and also passes through (4, 3).

🎯 Exam Tip: Pay attention to the coefficients when forming the perpendicular line equation. If the original line is \( Ax + By + C = 0 \), a perpendicular line can be written as \( Bx - Ay + k = 0 \).

Question 3. Find the equation of the straight line which passes through
(i) the origin and the point of intersection of the st. lines \( y - x + 7 = 0, y + 2x - 2 = 0 \);
(ii) the point \( (2, -9) \) and the intersection of the lines \( 2x + 5y - 8 = 0 \) and \( 3x - 4y = 35 \);
(iii) the origin and the point of intersection of the lines \( ax + by + c = 0 \) and \( a'x + b'y + c' = 0 \).
Answer:
(i) To find the equation of the line passing through the origin and the intersection of \( y - x + 7 = 0 \) and \( y + 2x - 2 = 0 \):
The given equations of the lines are:
\( y - x + 7 = 0 \) ...(1)
\( y + 2x - 2 = 0 \) ...(2)
The equation of a line that passes through the point of intersection of lines (1) and (2) is given by:
\( (y - x + 7) + k(y + 2x - 2) = 0 \) ...(3)
Since equation (3) passes through the origin (0, 0), we substitute x = 0 and y = 0 into it:
\( (0 - 0 + 7) + k(0 + 0 - 2) = 0 \)
\( 7 - 2k = 0 \)
\( 2k = 7 \)
\( k = \frac{7}{2} \)
Now, we put the value of \( k = \frac{7}{2} \) back into equation (3):
\( (y - x + 7) + \frac{7}{2}(y + 2x - 2) = 0 \)
To remove the fraction, multiply the entire equation by 2:
\( 2(y - x + 7) + 7(y + 2x - 2) = 0 \)
\( 2y - 2x + 14 + 7y + 14x - 14 = 0 \)
\( 12x + 9y = 0 \)
Divide the equation by 3:
\( 4x + 3y = 0 \)
This is the required equation of the line. This method is efficient for finding a line through an intersection.

(ii) To find the equation of the line passing through the point (2, -9) and the intersection of \( 2x + 5y - 8 = 0 \) and \( 3x - 4y = 35 \):
The given equations of the lines are:
\( 2x + 5y - 8 = 0 \) ...(1)
\( 3x - 4y - 35 = 0 \) ...(2) (Rewriting \( 3x - 4y = 35 \))
The equation of a line that passes through the point of intersection of lines (1) and (2) is given by:
\( (2x + 5y - 8) + k(3x - 4y - 35) = 0 \) ...(3)
Since equation (3) passes through the point (2, -9), we substitute x = 2 and y = -9 into it:
\( (2(2) + 5(-9) - 8) + k(3(2) - 4(-9) - 35) = 0 \)
\( (4 - 45 - 8) + k(6 + 36 - 35) = 0 \)
\( (-49) + k(7) = 0 \)
\( -49 + 7k = 0 \)
\( 7k = 49 \)
\( k = 7 \)
Now, we put the value of \( k = 7 \) back into equation (3):
\( (2x + 5y - 8) + 7(3x - 4y - 35) = 0 \)
\( 2x + 5y - 8 + 21x - 28y - 245 = 0 \)
\( 23x - 23y - 253 = 0 \)
Divide the entire equation by 23:
\( x - y - 11 = 0 \)
This is the required equation of the line. The value of k determines which specific line passes through the intersection.

(iii) To find the equation of the line passing through the origin and the intersection of \( ax + by + c = 0 \) and \( a'x + b'y + c' = 0 \):
The given equations of the lines are:
\( ax + by + c = 0 \) ...(1)
\( a'x + b'y + c' = 0 \) ...(2)
The equation of a line that passes through the point of intersection of lines (1) and (2) is given by:
\( (ax + by + c) + k(a'x + b'y + c') = 0 \) ...(3)
Since equation (3) passes through the origin (0, 0), we substitute x = 0 and y = 0 into it:
\( (a(0) + b(0) + c) + k(a'(0) + b'(0) + c') = 0 \)
\( c + kc' = 0 \)
\( kc' = -c \)
\( k = -\frac{c}{c'} \)
Now, we put the value of \( k = -\frac{c}{c'} \) back into equation (3):
\( (ax + by + c) - \frac{c}{c'}(a'x + b'y + c') = 0 \)
To remove the fraction, multiply the entire equation by \( c' \):
\( c'(ax + by + c) - c(a'x + b'y + c') = 0 \)
\( ac'x + bc'y + cc' - a'cx - b'cy - cc' = 0 \)
Group the terms with x and y:
\( (ac' - a'c)x + (bc' - b'c)y = 0 \)
This is the required equation of the straight line. This general form holds true for any two intersecting lines and the origin.
In simple words: We use a special formula for a line that goes through where two other lines cross. For each part, we use the given point (like the origin or (2, -9)) to find a specific number that makes the equation correct. This helps us find the exact straight line we need.

🎯 Exam Tip: The equation \( L_1 + kL_2 = 0 \) is a powerful tool for lines passing through the intersection of two lines. For lines passing through the origin, the constant term in the simplified equation will always be zero.

Question 4. Show that the equation \( n(ax + by + c) = c(lx + my + n) \) represents the line joining the origin to the point of intersection of \( ax + by + c = 0 \) and \( lx + my + n = 0 \).
Answer:
To show that the given equation represents the line joining the origin to the intersection of two other lines:
The given equations of the lines are:
\( ax + by + c = 0 \) ...(1)
\( lx + my + n = 0 \) ...(2)
The equation of a line that passes through the point of intersection of lines (1) and (2) is given by:
\( (ax + by + c) + k(lx + my + n) = 0 \) ...(3)
For this line to join the origin to the point of intersection, it must pass through the origin (0, 0).
So, we substitute x = 0 and y = 0 into equation (3):
\( (a(0) + b(0) + c) + k(l(0) + m(0) + n) = 0 \)
\( c + kn = 0 \)
\( kn = -c \)
\( k = -\frac{c}{n} \)
Now, we put the value of \( k = -\frac{c}{n} \) back into equation (3):
\( (ax + by + c) - \frac{c}{n}(lx + my + n) = 0 \)
To remove the fraction, multiply the entire equation by n:
\( n(ax + by + c) - c(lx + my + n) = 0 \)
Rearrange the terms to match the required form:
\( n(ax + by + c) = c(lx + my + n) \)
This matches the given equation, thus proving the statement. This method shows how a specific condition (passing through the origin) helps find the value of k.
In simple words: We are proving that a certain equation describes a straight line. This line starts at the point (0, 0) and goes exactly through the spot where two other given lines cross each other.

🎯 Exam Tip: When asked to "show that" an equation represents a certain line, typically you'll start with a general form (like \( L_1 + kL_2 = 0 \)) and use the given conditions (e.g., passing through the origin) to find the value of \( k \) that transforms it into the target equation.

Question 5. Find the equation of the line through the intersection of \( x - y = 1 \) and \( 2x - 3y + 1 = 0 \) and parallel to \( 3x + 4y = 12 \).
Answer:
To find the equation of a line passing through the intersection of two lines and parallel to a third line:
The given equations of the lines are:
\( x - y - 1 = 0 \) ...(1) (Rewriting \( x - y = 1 \))
\( 2x - 3y + 1 = 0 \) ...(2)
The equation of a line that passes through the point of intersection of lines (1) and (2) is given by:
\( (x - y - 1) + k(2x - 3y + 1) = 0 \) ...(3)
We can rearrange equation (3) to find its slope:
\( x + 2kx - y - 3ky - 1 + k = 0 \)
\( (1 + 2k)x + (-1 - 3k)y + (-1 + k) = 0 \)
The slope of this line (let's call it \( m_1 \)) is given by \( -\frac{\text{coefficient of x}}{\text{coefficient of y}} \):
\( m_1 = -\frac{(1 + 2k)}{(-1 - 3k)} = \frac{(1 + 2k)}{(1 + 3k)} \)
The third given line is \( 3x + 4y = 12 \), which can be written as \( 3x + 4y - 12 = 0 \).
The slope of this line (let's call it \( m_2 \)) is \( -\frac{3}{4} \).
Since the required line is parallel to \( 3x + 4y = 12 \), their slopes must be equal:
\( m_1 = m_2 \)
\( \frac{(1 + 2k)}{(1 + 3k)} = -\frac{3}{4} \)
Now, we cross-multiply to solve for k:
\( 4(1 + 2k) = -3(1 + 3k) \)
\( 4 + 8k = -3 - 9k \)
\( 8k + 9k = -3 - 4 \)
\( 17k = -7 \)
\( k = -\frac{7}{17} \)
Now, substitute the value of \( k \) back into the general equation of the line (equation 3) in the form \( (1 + 2k)x + (-1 - 3k)y + (-1 + k) = 0 \):
\( \left(1 + 2\left(-\frac{7}{17}\right)\right)x + \left(-1 - 3\left(-\frac{7}{17}\right)\right)y + \left(-1 + \left(-\frac{7}{17}\right)\right) = 0 \)
\( \left(1 - \frac{14}{17}\right)x + \left(-1 + \frac{21}{17}\right)y + \left(-1 - \frac{7}{17}\right) = 0 \)
\( \left(\frac{17 - 14}{17}\right)x + \left(\frac{-17 + 21}{17}\right)y + \left(\frac{-17 - 7}{17}\right) = 0 \)
\( \frac{3}{17}x + \frac{4}{17}y - \frac{24}{17} = 0 \)
Multiply the entire equation by 17 to clear the denominators:
\( 3x + 4y - 24 = 0 \)
This is the required equation of the line. The final equation of the line is a specific case found by satisfying both conditions.
In simple words: We need to find a line that passes through the crossing point of two other lines. This new line must also run in the exact same direction as a third given line. We use their slopes to figure out the missing number in our equation.

🎯 Exam Tip: When a line passes through the intersection of \( L_1 = 0 \) and \( L_2 = 0 \), its equation is \( L_1 + kL_2 = 0 \). If it's parallel to a third line, their slopes are equal. If it's perpendicular, the product of their slopes is -1. Use this logic consistently.

Question 6. Find the equation of the line through the intersection of \( x + 2y + 3 = 0 \) and \( 3x + 4y + 7 = 0 \) and parallel to \( y - x = 8 \).
Answer:
To find the equation of a line passing through the intersection of two lines and parallel to a third line:
The given equations of the lines are:
\( x + 2y + 3 = 0 \) ...(1)
\( 3x + 4y + 7 = 0 \) ...(2)
The equation of a line that passes through the point of intersection of lines (1) and (2) is given by:
\( (x + 2y + 3) + k(3x + 4y + 7) = 0 \) ...(3)
We rearrange equation (3) to find its slope:
\( x + 3kx + 2y + 4ky + 3 + 7k = 0 \)
\( (1 + 3k)x + (2 + 4k)y + (3 + 7k) = 0 \)
The slope of this line (let's call it \( m_1 \)) is given by \( -\frac{\text{coefficient of x}}{\text{coefficient of y}} \):
\( m_1 = -\frac{(1 + 3k)}{(2 + 4k)} \)
The third given line is \( y - x = 8 \), which can be written as \( x - y + 8 = 0 \) or \( y = x + 8 \).
The slope of this line (let's call it \( m_2 \)) is \( 1 \).
Since the required line is parallel to \( y - x = 8 \), their slopes must be equal:
\( m_1 = m_2 \)
\( -\frac{(1 + 3k)}{(2 + 4k)} = 1 \)
Now, we cross-multiply to solve for k:
\( -(1 + 3k) = (2 + 4k) \)
\( -1 - 3k = 2 + 4k \)
\( -1 - 2 = 4k + 3k \)
\( -3 = 7k \)
\( k = -\frac{3}{7} \)
Now, substitute the value of \( k \) back into the general equation of the line (equation 3) in the form \( (1 + 3k)x + (2 + 4k)y + (3 + 7k) = 0 \):
\( \left(1 + 3\left(-\frac{3}{7}\right)\right)x + \left(2 + 4\left(-\frac{3}{7}\right)\right)y + \left(3 + 7\left(-\frac{3}{7}\right)\right) = 0 \)
\( \left(1 - \frac{9}{7}\right)x + \left(2 - \frac{12}{7}\right)y + \left(3 - 3\right) = 0 \)
\( \left(\frac{7 - 9}{7}\right)x + \left(\frac{14 - 12}{7}\right)y + 0 = 0 \)
\( -\frac{2}{7}x + \frac{2}{7}y = 0 \)
Multiply the entire equation by \( -\frac{7}{2} \) to simplify:
\( x - y = 0 \)
This is the required equation of the line. This shows that the line passes through the origin because its constant term is zero.
In simple words: We need to find a line that goes through the meeting point of the first two lines. This new line must also travel in the same direction as the third line. We use the idea of slopes being equal for parallel lines to solve this.

🎯 Exam Tip: When a line is parallel to \( y - x = 8 \), its slope must be 1. Be careful with calculations involving fractions to ensure accuracy, especially when substituting the value of k back into the combined equation.

Question 7. Find the equation of the line through the intersection of \( y + x = 0 \) and \( 2x - 3y + 7 = 0 \), and perpendicular to the line \( 2y - 3x - 5 = 0 \).
Answer:
To find the equation of a line passing through the intersection of two lines and perpendicular to a third line:
The given equations of the lines are:
\( x + y = 0 \) ...(1) (Rewriting \( y + x = 0 \))
\( 2x - 3y + 7 = 0 \) ...(2)
The equation of a line that passes through the point of intersection of lines (1) and (2) is given by:
\( (x + y) + k(2x - 3y + 7) = 0 \) ...(3)
We rearrange equation (3) to find its slope:
\( x + 2kx + y - 3ky + 7k = 0 \)
\( (1 + 2k)x + (1 - 3k)y + 7k = 0 \)
The slope of this line (let's call it \( m_1 \)) is given by \( -\frac{\text{coefficient of x}}{\text{coefficient of y}} \):
\( m_1 = -\frac{(1 + 2k)}{(1 - 3k)} \)
The third given line is \( 2y - 3x - 5 = 0 \), which can be written as \( 2y = 3x + 5 \), or \( y = \frac{3}{2}x + \frac{5}{2} \).
The slope of this line (let's call it \( m_2 \)) is \( \frac{3}{2} \).
Since the required line is perpendicular to \( 2y - 3x - 5 = 0 \), the product of their slopes must be -1:
\( m_1 \cdot m_2 = -1 \)
\( -\frac{(1 + 2k)}{(1 - 3k)} \cdot \frac{3}{2} = -1 \)
\( -\frac{3(1 + 2k)}{2(1 - 3k)} = -1 \)
Multiply both sides by -1:
\( \frac{3(1 + 2k)}{2(1 - 3k)} = 1 \)
Now, we cross-multiply to solve for k:
\( 3(1 + 2k) = 2(1 - 3k) \)
\( 3 + 6k = 2 - 6k \)
\( 6k + 6k = 2 - 3 \)
\( 12k = -1 \)
\( k = -\frac{1}{12} \)
Now, substitute the value of \( k \) back into the general equation of the line (equation 3) in the form \( (1 + 2k)x + (1 - 3k)y + 7k = 0 \):
\( \left(1 + 2\left(-\frac{1}{12}\right)\right)x + \left(1 - 3\left(-\frac{1}{12}\right)\right)y + 7\left(-\frac{1}{12}\right) = 0 \)
\( \left(1 - \frac{1}{6}\right)x + \left(1 + \frac{1}{4}\right)y - \frac{7}{12} = 0 \)
\( \left(\frac{6 - 1}{6}\right)x + \left(\frac{4 + 1}{4}\right)y - \frac{7}{12} = 0 \)
\( \frac{5}{6}x + \frac{5}{4}y - \frac{7}{12} = 0 \)
Multiply the entire equation by 12 (the least common multiple of 6, 4, and 12) to clear the denominators:
\( 12\left(\frac{5}{6}x\right) + 12\left(\frac{5}{4}y\right) - 12\left(\frac{7}{12}\right) = 0 \)
\( 10x + 15y - 7 = 0 \)
This is the required equation of the line. This problem combines the concept of lines through intersection and perpendicularity effectively.
In simple words: We need to find a line that goes through the point where the first two lines cross. This new line must also be at a right angle (90 degrees) to a third given line. We use the rule that slopes of perpendicular lines multiply to -1.

🎯 Exam Tip: Remember that for perpendicular lines, if one slope is \( m \), the other is \( -\frac{1}{m} \). Pay careful attention to algebraic simplification, especially when multiplying by the LCM to clear fractions.