Samacheer Kalvi Class 10 Maths Solutions Chapter 5 Coordinate Geometry Exercise 5.4

Question 1. Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x - \( \frac { 3 }{ 17 } \) = 0
Answer:
(i) For the line 5y – 3 = 0:
We can rewrite this as 5y = 3.
Dividing by 5, we get y = \( \frac { 3 }{ 5 } \).
This equation is in the form y = c, which represents a horizontal line. A horizontal line has a slope of 0. This is because there is no change in y as x changes.
Thus, Slope = 0.

(ii) For the line 7x - \( \frac { 3 }{ 17 } \) = 0:
We can rewrite this as 7x = \( \frac { 3 }{ 17 } \).
Dividing by 7, we get x = \( \frac { 3 }{ 17 \times 7 } \) = \( \frac { 3 }{ 119 } \).
This equation is in the form x = c, which represents a vertical line. A vertical line runs straight up and down and is parallel to the Y-axis.
Thus, Slope is undefined.
In simple words: For a horizontal line (y = constant), the slope is zero. For a vertical line (x = constant), the slope is undefined because it goes straight up or down.

🎯 Exam Tip: Remember that the slope of a horizontal line is 0, and the slope of a vertical line is undefined. Visualize these lines to easily recall their slopes.

Question 2. Find the slope of the line which is
(i) parallel to y = 0.7x – 11
(ii) perpendicular to the line x = -11
Answer:
(i) We are looking for the slope of a line parallel to y = 0.7x - 11.
This equation is already in the slope-intercept form y = mx + c, where 'm' is the slope and 'c' is the y-intercept. The slope of this line is 0.7.
Lines that are parallel to each other always have the same slope. Therefore, the slope of the required line is 0.7.

(ii) We need the slope of a line perpendicular to x = -11.
The equation x = -11 represents a vertical line. A vertical line has an undefined slope because it is parallel to the Y-axis.
A line perpendicular to a vertical line must be a horizontal line. A horizontal line has a slope of 0. Thus, the slope of the required line is 0.
In simple words: Parallel lines have the same slope. Perpendicular lines have slopes that are negative reciprocals of each other. If one line is vertical (slope undefined), the line perpendicular to it will be horizontal (slope 0).

🎯 Exam Tip: Parallel lines share the exact same slope. Perpendicular lines have slopes where one is the negative inverse of the other; if \( m_1 \) is the slope of the first line, the slope of the perpendicular line \( m_2 \) is \( -\frac{1}{m_1} \). This relationship doesn't apply when one of the lines is perfectly vertical or horizontal, as their slopes are undefined or zero respectively. In such cases, remember that a vertical line has an undefined slope, and a horizontal line has a slope of zero.

Question 3. Check whether the given lines are parallel or perpendicular
(i) \( \frac { x }{ 3 } + \frac { y }{ 4 } + \frac { 1 }{ 7 } = 0 \) and \( \frac { 2x }{ 3 } + \frac { y }{ 2 } + \frac { 1 }{ 10 } = 0 \)
(ii) 5x + 23y + 14 = 0 and 23x – 5y + 9 = 0
Answer:
We use the formula for the slope of a line \( Ax + By + C = 0 \), which is \( m = -\frac{A}{B} \).

(i) For the first line: \( \frac { x }{ 3 } + \frac { y }{ 4 } + \frac { 1 }{ 7 } = 0 \)
Here, A = \( \frac { 1 }{ 3 } \) and B = \( \frac { 1 }{ 4 } \).
Slope \( m_1 = -\frac{A}{B} = - \frac { 1/3 }{ 1/4 } = - \frac { 1 }{ 3 } \times \frac { 4 }{ 1 } = - \frac { 4 }{ 3 } \)
For the second line: \( \frac { 2x }{ 3 } + \frac { y }{ 2 } + \frac { 1 }{ 10 } = 0 \)
Here, A = \( \frac { 2 }{ 3 } \) and B = \( \frac { 1 }{ 2 } \).
Slope \( m_2 = -\frac{A}{B} = - \frac { 2/3 }{ 1/2 } = - \frac { 2 }{ 3 } \times \frac { 2 }{ 1 } = - \frac { 4 }{ 3 } \)
Since \( m_1 = m_2 = - \frac { 4 }{ 3 } \), the two lines are parallel. Parallel lines have the same slope, meaning they will never intersect.

(ii) For the first line: 5x + 23y + 14 = 0
Here, A = 5 and B = 23.
Slope \( m_1 = -\frac{A}{B} = - \frac { 5 }{ 23 } \)
For the second line: 23x – 5y + 9 = 0
Here, A = 23 and B = -5.
Slope \( m_2 = -\frac{A}{B} = - \frac { 23 }{ -5 } = \frac { 23 }{ 5 } \)
Now, let's check the product of the slopes:
\( m_1 \times m_2 = \left( - \frac { 5 }{ 23 } \right) \times \left( \frac { 23 }{ 5 } \right) = -1 \)
Since the product of their slopes is -1, the two lines are perpendicular. This means they intersect at a 90-degree angle.
In simple words: Two lines are parallel if their slopes are exactly the same. They are perpendicular if you multiply their slopes and get -1.

🎯 Exam Tip: When given equations in the form \( Ax + By + C = 0 \), quickly find the slope using \( m = -\frac{A}{B} \). Then, compare slopes: \( m_1 = m_2 \) for parallel, and \( m_1 m_2 = -1 \) for perpendicular.

Question 4. If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find 'p'
Answer:
For the first line: 12y = -(p + 3)x + 12
To find the slope, we rearrange it into the y = mx + c form:
y = \( -\frac{(p+3)}{12}x + \frac{12}{12} \)
y = \( -\frac{(p+3)}{12}x + 1 \)
So, the slope of the first line, \( m_1 = -\frac{(p+3)}{12} \).

For the second line: 12x – 7y = 16
This is in the form Ax + By + C = 0. The slope is \( m = -\frac{A}{B} \).
Here, A = 12 and B = -7.
So, the slope of the second line, \( m_2 = -\frac{12}{-7} = \frac{12}{7} \).

The problem states that the two lines are perpendicular. For perpendicular lines, the product of their slopes is -1.
\( m_1 \times m_2 = -1 \)
\( -\frac{(p+3)}{12} \times \frac{12}{7} = -1 \)
We can cancel the 12s.
\( -\frac{(p+3)}{7} = -1 \)
Multiply both sides by -1 to remove the negative signs.
\( \frac{(p+3)}{7} = 1 \)
Multiply both sides by 7.
\( p+3 = 7 \)
Subtract 3 from both sides.
\( p = 7 - 3 \)
\( p = 4 \)
The value of p is 4.
In simple words: We found the slope of each line using the given equations. Since the lines are perpendicular, we set the product of their slopes equal to -1 and then solved the equation to find the value of 'p'.

🎯 Exam Tip: Always convert line equations to either slope-intercept form (y = mx + c) or general form (Ax + By + C = 0) to easily identify their slopes. For perpendicular lines, \( m_1m_2 = -1 \); for parallel lines, \( m_1 = m_2 \).

Question 5. Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
Answer:
First, we need to find the slope of the line joining points Q(3, -2) and R(-5,4). We'll call this slope \( m_{QR} \).
The formula for the slope between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
For Q(3, -2) and R(-5, 4):
\( m_{QR} = \frac{4 - (-2)}{-5 - 3} = \frac{4 + 2}{-8} = \frac{6}{-8} = -\frac{3}{4} \)
The line we need to find is parallel to QR. Parallel lines have the same slope.
So, the slope of our required line, m, is also \( -\frac{3}{4} \).
The required line passes through point P(-5, 2) and has a slope of \( -\frac{3}{4} \). We use the point-slope form of a linear equation: \( y - y_1 = m(x - x_1) \).
Here, \( (x_1, y_1) = (-5, 2) \) and \( m = -\frac{3}{4} \).
\( y - 2 = -\frac{3}{4}(x - (-5)) \)
\( y - 2 = -\frac{3}{4}(x + 5) \)
Multiply both sides by 4 to clear the fraction.
\( 4(y - 2) = -3(x + 5) \)
\( 4y - 8 = -3x - 15 \)
Now, rearrange the equation into the general form \( Ax + By + C = 0 \).
\( 3x + 4y - 8 + 15 = 0 \)
\( 3x + 4y + 7 = 0 \)
The equation of the straight line is \( 3x + 4y + 7 = 0 \). This line passes through P and is parallel to the line QR.
In simple words: First, find the slope of the line connecting Q and R. Since our new line is parallel, it has the same slope. Then, use this slope and point P to write the equation of the new line.

🎯 Exam Tip: When a line is parallel to another, their slopes are identical. If you're given two points, always use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) correctly. Then, the point-slope form \( y - y_1 = m(x - x_1) \) is often the quickest way to find the equation of the line.

Question 6. Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Answer:
Let the given point be D(6, -2). Let the two points forming the line segment be A(6, 7) and B(2, -3).
First, we find the slope of the line joining points A(6, 7) and B(2, -3). Let's call this slope \( m_{AB} \).
Using the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \):
\( m_{AB} = \frac{-3 - 7}{2 - 6} = \frac{-10}{-4} = \frac{5}{2} \)
The line we need to find is perpendicular to line AB. If two lines are perpendicular, the product of their slopes is -1. So, if \( m_{AB} \) is the slope of AB, the slope of the perpendicular line, m, will be \( -\frac{1}{m_{AB}} \).
\( m = -\frac{1}{5/2} = -\frac{2}{5} \)
Now, we have the slope \( m = -\frac{2}{5} \) and the point D(6, -2) through which the line passes. We use the point-slope form: \( y - y_1 = m(x - x_1) \).
Here, \( (x_1, y_1) = (6, -2) \) and \( m = -\frac{2}{5} \).
\( y - (-2) = -\frac{2}{5}(x - 6) \)
\( y + 2 = -\frac{2}{5}(x - 6) \)
Multiply both sides by 5 to remove the fraction.
\( 5(y + 2) = -2(x - 6) \)
\( 5y + 10 = -2x + 12 \)
Rearrange into the general form \( Ax + By + C = 0 \).
\( 2x + 5y + 10 - 12 = 0 \)
\( 2x + 5y - 2 = 0 \)
The equation of the straight line is \( 2x + 5y - 2 = 0 \). This line passes through D and is perpendicular to the line segment AB.
In simple words: First, calculate the slope of the line connecting points A and B. Since our new line needs to be perpendicular, its slope will be the negative inverse of the first slope. Then, use this new slope and the given point D to find the equation of the line.

🎯 Exam Tip: Remember that the slope of a perpendicular line is the negative reciprocal of the original line's slope. Be careful with signs when calculating slopes and cross-multiplying during equation formation.

Question 7. A(-3,0) B(10, -2) and C(12,3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Answer:
An altitude of a triangle is a line segment from a vertex to the opposite side, forming a right angle with that side.
The vertices of ∆ABC are A(-3, 0), B(10, -2), and C(12, 3).

**1. Equation of the altitude from A (AD):**
This altitude is perpendicular to side BC and passes through point A(-3, 0).
First, find the slope of BC. For B(10, -2) and C(12, 3):
\( m_{BC} = \frac{3 - (-2)}{12 - 10} = \frac{3 + 2}{2} = \frac{5}{2} \)
Since the altitude AD is perpendicular to BC, its slope \( m_{AD} \) will be the negative reciprocal of \( m_{BC} \).
\( m_{AD} = -\frac{1}{m_{BC}} = -\frac{1}{5/2} = -\frac{2}{5} \)
Now, use the point-slope form for the altitude AD, passing through A(-3, 0) with slope \( -\frac{2}{5} \):
\( y - y_1 = m(x - x_1) \)
\( y - 0 = -\frac{2}{5}(x - (-3)) \)
\( y = -\frac{2}{5}(x + 3) \)
Multiply by 5:
\( 5y = -2(x + 3) \)
\( 5y = -2x - 6 \)
Rearrange into general form:
\( 2x + 5y + 6 = 0 \)
This is the equation of the altitude from A.

**2. Equation of the altitude from B (BE):**
This altitude is perpendicular to side AC and passes through point B(10, -2).
First, find the slope of AC. For A(-3, 0) and C(12, 3):
\( m_{AC} = \frac{3 - 0}{12 - (-3)} = \frac{3}{12 + 3} = \frac{3}{15} = \frac{1}{5} \)
Since the altitude BE is perpendicular to AC, its slope \( m_{BE} \) will be the negative reciprocal of \( m_{AC} \).
\( m_{BE} = -\frac{1}{m_{AC}} = -\frac{1}{1/5} = -5 \)
Now, use the point-slope form for the altitude BE, passing through B(10, -2) with slope -5:
\( y - y_1 = m(x - x_1) \)
\( y - (-2) = -5(x - 10) \)
\( y + 2 = -5x + 50 \)
Rearrange into general form:
\( 5x + y + 2 - 50 = 0 \)
\( 5x + y - 48 = 0 \)
This is the equation of the altitude from B.

Both altitudes meet at a point called the orthocenter, a key feature of a triangle.
In simple words: To find an altitude's equation, first calculate the slope of the opposite side. Then, find the negative inverse of that slope for the altitude. Finally, use the altitude's slope and the vertex it comes from to write its equation. Do this for both vertex A and vertex B.

🎯 Exam Tip: Clearly identify the side opposite to the vertex from which the altitude is drawn. Remember the relationship between slopes of perpendicular lines. Work systematically for each altitude to avoid errors.

Question 8. Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B (6,-4).
Answer:
A perpendicular bisector is a line that cuts another line segment into two equal halves (bisects it) and is also perpendicular to it.

**1. Find the midpoint of AB:**
Let A = \( (x_1, y_1) = (-4, 2) \) and B = \( (x_2, y_2) = (6, -4) \).
The midpoint formula is \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
Midpoint \( C = \left( \frac{-4 + 6}{2}, \frac{2 + (-4)}{2} \right) \)
\( C = \left( \frac{2}{2}, \frac{-2}{2} \right) \)
\( C = (1, -1) \)

**2. Find the slope of AB:**
Slope \( m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \)
\( m_{AB} = \frac{-4 - 2}{6 - (-4)} = \frac{-6}{6 + 4} = \frac{-6}{10} = -\frac{3}{5} \)

**3. Find the slope of the perpendicular bisector:**
The perpendicular bisector's slope (\( m_{\perp} \)) is the negative reciprocal of \( m_{AB} \).
\( m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3} \)

**4. Find the equation of the perpendicular bisector:**
Use the point-slope form \( y - y_1 = m(x - x_1) \) with the midpoint C(1, -1) and slope \( m_{\perp} = \frac{5}{3} \).
\( y - (-1) = \frac{5}{3}(x - 1) \)
\( y + 1 = \frac{5}{3}(x - 1) \)
Multiply both sides by 3 to eliminate the fraction.
\( 3(y + 1) = 5(x - 1) \)
\( 3y + 3 = 5x - 5 \)
Rearrange into the general form \( Ax + By + C = 0 \).
\( 5x - 3y - 5 - 3 = 0 \)
\( 5x - 3y - 8 = 0 \)
The equation of the perpendicular bisector is \( 5x - 3y - 8 = 0 \). This line cuts the segment AB exactly in half and at a 90-degree angle.
In simple words: First, find the middle point of the line segment A and B. Next, calculate the slope of the line AB. Then, find the slope that is perpendicular to AB. Finally, use the middle point and the perpendicular slope to write the equation of the line.

🎯 Exam Tip: Two key steps for perpendicular bisectors: find the midpoint (average of coordinates) and the negative reciprocal of the slope. Combine these into the point-slope form of the line equation.

Question 9. Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Answer:
**1. Find the point of intersection of the first two lines:**
Given lines:
(1) \( 7x + 3y = 10 \)
(2) \( 5x - 4y = 1 \)
We use the elimination method to solve for x and y.
Multiply equation (1) by 4: \( 28x + 12y = 40 \) ... (3)
Multiply equation (2) by 3: \( 15x - 12y = 3 \) ... (4)
Add equation (3) and (4):
\( (28x + 12y) + (15x - 12y) = 40 + 3 \)
\( 43x = 43 \)
\( x = \frac{43}{43} = 1 \)
Substitute the value of x = 1 into equation (1):
\( 7(1) + 3y = 10 \)
\( 7 + 3y = 10 \)
\( 3y = 10 - 7 \)
\( 3y = 3 \)
\( y = \frac{3}{3} = 1 \)
So, the point of intersection is (1, 1).

**2. Find the slope of the line parallel to the required line:**
The line is \( 13x + 5y + 12 = 0 \). This is in the form \( Ax + By + C = 0 \).
Its slope is \( m = -\frac{A}{B} = -\frac{13}{5} \).
Since our required line is parallel to this line, its slope will also be \( m = -\frac{13}{5} \).

**3. Find the equation of the required line:**
The required line passes through the point of intersection (1, 1) and has a slope of \( -\frac{13}{5} \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 1 = -\frac{13}{5}(x - 1) \)
Multiply both sides by 5:
\( 5(y - 1) = -13(x - 1) \)
\( 5y - 5 = -13x + 13 \)
Rearrange into the general form \( Ax + By + C = 0 \):
\( 13x + 5y - 5 - 13 = 0 \)
\( 13x + 5y - 18 = 0 \)
The equation of the straight line is \( 13x + 5y - 18 = 0 \).
In simple words: First, solve the first two line equations to find where they cross. That point is (1,1). Next, find the slope of the third line. Since our new line is parallel, it will have the same slope. Finally, use this slope and the intersection point to write the equation of the new line.

🎯 Exam Tip: For problems involving the intersection of lines, the elimination or substitution method for solving simultaneous equations is crucial. Remember that parallel lines share the same slope, a fundamental property in coordinate geometry.

Question 10. Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
Answer:
**1. Find the point of intersection of the first two lines:**
Given lines:
(1) \( 5x - 6y = 2 \)
(2) \( 3x + 2y = 10 \)
We use the elimination method.
Multiply equation (2) by 3 to make the y-coefficients opposites:
\( 3 \times (3x + 2y = 10) \implies 9x + 6y = 30 \) ... (3)
Now, add equation (1) and (3):
\( (5x - 6y) + (9x + 6y) = 2 + 30 \)
\( 14x = 32 \)
\( x = \frac{32}{14} = \frac{16}{7} \)
Substitute the value of \( x = \frac{16}{7} \) into equation (2):
\( 3\left(\frac{16}{7}\right) + 2y = 10 \)
\( \frac{48}{7} + 2y = 10 \)
\( 2y = 10 - \frac{48}{7} \)
To subtract, find a common denominator:
\( 2y = \frac{70}{7} - \frac{48}{7} \)
\( 2y = \frac{22}{7} \)
Divide by 2:
\( y = \frac{22}{7 \times 2} = \frac{11}{7} \)
So, the point of intersection is \( \left(\frac{16}{7}, \frac{11}{7}\right) \).

**2. Find the slope of the line perpendicular to the required line:**
The line is \( 4x - 7y + 13 = 0 \). This is in the form \( Ax + By + C = 0 \).
Its slope is \( m_{given} = -\frac{A}{B} = -\frac{4}{-7} = \frac{4}{7} \).
Since our required line is perpendicular to this line, its slope \( m_{required} \) will be the negative reciprocal of \( m_{given} \).
\( m_{required} = -\frac{1}{m_{given}} = -\frac{1}{4/7} = -\frac{7}{4} \).

**3. Find the equation of the required line:**
The required line passes through the point of intersection \( \left(\frac{16}{7}, \frac{11}{7}\right) \) and has a slope of \( -\frac{7}{4} \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - \frac{11}{7} = -\frac{7}{4}\left(x - \frac{16}{7}\right) \)
To remove fractions, multiply by the least common multiple of the denominators (7 and 4), which is 28.
\( 28\left(y - \frac{11}{7}\right) = 28\left(-\frac{7}{4}\left(x - \frac{16}{7}\right)\right) \)
\( 28y - 44 = -49\left(x - \frac{16}{7}\right) \)
\( 28y - 44 = -49x + 49\left(\frac{16}{7}\right) \)
\( 28y - 44 = -49x + 7 \times 16 \)
\( 28y - 44 = -49x + 112 \)
Rearrange into the general form \( Ax + By + C = 0 \):
\( 49x + 28y - 44 - 112 = 0 \)
\( 49x + 28y - 156 = 0 \)
The equation of the straight line is \( 49x + 28y - 156 = 0 \).
In simple words: First, find the exact point where the first two lines meet. This point will be on our new line. Next, figure out the slope of the third line. Since our new line needs to be perpendicular to it, we find the negative inverse of that slope. Finally, use this new slope and the intersection point to write the equation of the required line.

🎯 Exam Tip: When dealing with fractions in calculations, it's often best to find a common denominator or multiply the entire equation by the LCM of denominators to simplify. Double-check all signs, especially when dealing with negative reciprocals for perpendicular lines.

Question 11. Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
Answer:
**1. Find the first point of intersection (P1):**
Given lines:
(A) \( 3x + y + 2 = 0 \implies 3x + y = -2 \)
(B) \( x - 2y - 4 = 0 \implies x - 2y = 4 \)
Multiply equation (A) by 2: \( 6x + 2y = -4 \) ... (A')
Add equation (A') and (B):
\( (6x + 2y) + (x - 2y) = -4 + 4 \)
\( 7x = 0 \)
\( x = 0 \)
Substitute \( x = 0 \) into equation (A):
\( 3(0) + y = -2 \)
\( y = -2 \)
So, the first point of intersection \( P_1 \) is (0, -2).

**2. Find the second point of intersection (P2):**
Given lines:
(C) \( 7x - 3y = -12 \)
(D) \( 2y = x + 3 \implies -x + 2y = 3 \)
Multiply equation (D) by 3: \( -3x + 6y = 9 \) ... (D')
Multiply equation (C) by 2: \( 14x - 6y = -24 \) ... (C')
Add equation (D') and (C'):
\( (-3x + 6y) + (14x - 6y) = 9 + (-24) \)
\( 11x = -15 \)
\( x = -\frac{15}{11} \)
Substitute \( x = -\frac{15}{11} \) into equation (D):
\( 2y = -\frac{15}{11} + 3 \)
\( 2y = -\frac{15}{11} + \frac{33}{11} \)
\( 2y = \frac{18}{11} \)
\( y = \frac{18}{11 \times 2} = \frac{9}{11} \)
So, the second point of intersection \( P_2 \) is \( \left(-\frac{15}{11}, \frac{9}{11}\right) \).

**3. Find the equation of the line joining P1(0, -2) and P2\( \left(-\frac{15}{11}, \frac{9}{11}\right) \):**
We use the two-point form of a linear equation: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \).
Let \( (x_1, y_1) = (0, -2) \) and \( (x_2, y_2) = \left(-\frac{15}{11}, \frac{9}{11}\right) \).
\( \frac{y - (-2)}{\frac{9}{11} - (-2)} = \frac{x - 0}{-\frac{15}{11} - 0} \)
\( \frac{y + 2}{\frac{9}{11} + \frac{22}{11}} = \frac{x}{-\frac{15}{11}} \)
\( \frac{y + 2}{\frac{31}{11}} = \frac{x}{-\frac{15}{11}} \)
Multiply both sides by 11 to simplify:
\( \frac{y + 2}{31} = \frac{x}{-15} \)
Cross-multiply:
\( -15(y + 2) = 31x \)
\( -15y - 30 = 31x \)
Rearrange into general form \( Ax + By + C = 0 \):
\( 31x + 15y + 30 = 0 \)
The equation of the straight line is \( 31x + 15y + 30 = 0 \).
In simple words: First, solve the first set of two line equations to find the first crossing point. Then, solve the second set of two line equations to find the second crossing point. Finally, use these two crossing points to find the equation of the straight line that connects them.

🎯 Exam Tip: This question requires solving two pairs of simultaneous equations and then using the two intersection points to find a new line's equation. Be meticulous with arithmetic, especially when dealing with fractions.

Question 12. Find the equation of a straight line through the point of intersection of the lines 8x + 3y = 18, 4x + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Answer:
**1. Find the point of intersection of the first two lines (P):**
Given lines:
(1) \( 8x + 3y = 18 \)
(2) \( 4x + 5y = 9 \)
Multiply equation (2) by 2:
\( 2 \times (4x + 5y = 9) \implies 8x + 10y = 18 \) ... (3)
Subtract equation (1) from equation (3):
\( (8x + 10y) - (8x + 3y) = 18 - 18 \)
\( 7y = 0 \)
\( y = 0 \)
Substitute \( y = 0 \) into equation (1):
\( 8x + 3(0) = 18 \)
\( 8x = 18 \)
\( x = \frac{18}{8} = \frac{9}{4} \)
So, the point of intersection P is \( \left(\frac{9}{4}, 0\right) \).

**2. Find the midpoint of the line segment joining (5, -4) and (-7, 6) (M):**
Let \( (x_1, y_1) = (5, -4) \) and \( (x_2, y_2) = (-7, 6) \).
Midpoint \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
\( M = \left( \frac{5 + (-7)}{2}, \frac{-4 + 6}{2} \right) \)
\( M = \left( \frac{-2}{2}, \frac{2}{2} \right) \)
\( M = (-1, 1) \)

**3. Find the equation of the line joining P\( \left(\frac{9}{4}, 0\right) \) and M(-1, 1):**
We use the two-point form: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \).
Let \( (x_1, y_1) = \left(\frac{9}{4}, 0\right) \) and \( (x_2, y_2) = (-1, 1) \).
\( \frac{y - 0}{1 - 0} = \frac{x - \frac{9}{4}}{-1 - \frac{9}{4}} \)
\( \frac{y}{1} = \frac{x - \frac{9}{4}}{-\frac{4}{4} - \frac{9}{4}} \)
\( y = \frac{x - \frac{9}{4}}{-\frac{13}{4}} \)
Multiply both sides by \( -\frac{13}{4} \):
\( -\frac{13}{4} y = x - \frac{9}{4} \)
Multiply the entire equation by 4 to remove fractions:
\( -13y = 4x - 9 \)
Rearrange into the general form \( Ax + By + C = 0 \):
\( 4x + 13y - 9 = 0 \)
The equation of the straight line is \( 4x + 13y - 9 = 0 \).
In simple words: First, find the point where the first two lines cross. Then, find the middle point of the given line segment. Finally, use these two special points to write the equation of the straight line connecting them.

🎯 Exam Tip: This problem combines finding an intersection point and a midpoint. Treat each step as distinct, focusing on accuracy for the coordinates and then using the two-point form correctly to derive the final line equation.