Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.9

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.9

Chapter 3

 

Question 1. Fill in the blanks:
(i) \( y = px \) where \( p \in Z \) always passes through the ______.
(ii) The intersecting point of the line \( x = 4 \) and \( y = -4 \) is ______.
(iii) Scale for the given graph,
On the x-axis 1 cm = ______ units
y-axis 1 cm = ______ units
Answer:
(i) Origin (0,0)
(ii) (4,-4)
(iii) 3 units, 25 units. This scale helps in accurately representing the data points and visualizing the linear relationships.
In simple words: The first part is asking about where the line of a certain equation always crosses. The second asks for the exact spot where two lines meet. The third part asks for the measurement scale used on a graph.

🎯 Exam Tip: For fill-in-the-blanks, always double-check the context and basic properties of equations and graphs to ensure your answers are precise.

 

Question 2.
(i) The points (1,1) (2,2) (3,3) lie on a same straight line.
(ii) \( y = -9x \) not passes through the origin.
Answer:
(i) True. These points satisfy the equation \( y = x \), which represents a straight line. Every point on this line has equal x and y coordinates.
(ii) False. If you put \( x = 0 \) into the equation \( y = -9x \), you get \( y = -9 \times 0 \), which means \( y = 0 \). So, the line passes through (0,0), which is the origin.
In simple words: Part (i) asks if points like (1,1), (2,2), (3,3) are on the same straight line, which is true because y equals x for all of them. Part (ii) asks if the line \( y = -9x \) does not go through the center of the graph (the origin); this is false, because it does.

🎯 Exam Tip: To check if points lie on a line, substitute their coordinates into the line's equation. If the equation holds true for all points, they lie on the line.

 

Question 3. Will a line pass through (2, 2) if it intersects the axes at (2, 0) and (0, 2)?
Answer:
A line that intersects the x-axis at (a, 0) and the y-axis at (0, b) can be written using the intercept form as \( \frac{x}{a} + \frac{y}{b} = 1 \). This form is useful when you know where a line crosses both axes.
Here, the intercepts are \( a = 2 \) and \( b = 2 \).
So, the equation of the line is \( \frac{x}{2} + \frac{y}{2} = 1 \).
This simplifies to \( x + y = 2 \).
Now, let's check if the point (2, 2) lies on this line. We substitute \( x = 2 \) and \( y = 2 \) into the equation:
\( 2 + 2 = 2 \)
\( 4 = 2 \)
This statement is false, as 4 is not equal to 2.
Since the point (2, 2) does not satisfy the equation of the line, the line does not pass through (2, 2).
In simple words: First, we find the equation of the line using the points where it crosses the x and y lines. Then, we check if the point (2,2) fits this equation. If it does not, then the line does not go through that point.

🎯 Exam Tip: When given x and y intercepts, use the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \) to quickly find the line's equation, then substitute the given point to verify if it lies on the line.

 

Question 4. A line passing through (4, – 2) and intersects the Y-axis at (0, 2). Find a point on the line in the second quadrant.
Answer:
We are given two points on the line: \( (x_1, y_1) = (4, -2) \) and \( (x_2, y_2) = (0, 2) \). We use the two-point formula to find the equation of the line: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \). This formula helps us define any straight line when we know two points it passes through.
Substitute the given points:
\( \frac{y - (-2)}{2 - (-2)} = \frac{x - 4}{0 - 4} \)
\( \frac{y + 2}{4} = \frac{x - 4}{-4} \)
Multiply both sides by 4 to simplify:
\( y + 2 = -(x - 4) \)
\( y + 2 = -x + 4 \)
Rearrange the terms to get the equation in standard form:
\( x + y = 4 - 2 \)
\( x + y = 2 \). This is the equation of the line.
A point in the second quadrant has a negative x-coordinate and a positive y-coordinate. Let's choose an x-value of \( -2 \).
Substitute \( x = -2 \) into the equation \( x + y = 2 \):
\( -2 + y = 2 \)
\( y = 2 + 2 \)
\( y = 4 \)
So, the point on the line in the second quadrant is \( (-2, 4) \).
In simple words: First, we find the rule (equation) for the line using the two points given. Then, we find a point on this line that is in the "second square" of the graph, which means its x-value is negative and its y-value is positive. We do this by choosing a negative x-value and solving for y.

🎯 Exam Tip: Remember the signs of coordinates in each quadrant: Q1 (+,+), Q2 (-,+), Q3 (-,-), Q4 (+,-). This helps in identifying or finding points in specific quadrants.

 

Question 5. If the points P(5, 3) Q(-3, 3) R (-3, -4) and S form a rectangle then find the coordinate of S.
Answer:
In a rectangle, opposite sides are parallel and equal in length. This property helps us find the missing coordinate. Given points are P(5, 3), Q(-3, 3), R(-3, -4).
Plotting the points approximately on a graph:
P and Q have the same y-coordinate (3), so PQ is a horizontal line.
Q and R have the same x-coordinate (-3), so QR is a vertical line.
Since PQ is horizontal, RS must also be horizontal and parallel to PQ. This means S will have the same y-coordinate as R, which is -4.
Since QR is vertical, PS must also be vertical and parallel to QR. This means S will have the same x-coordinate as P, which is 5.
Therefore, the coordinates of point S are (5, -4).

In simple words: For a rectangle, opposite points share either the same x-value or the same y-value. By looking at points P and R, we can find the x-coordinate of S from P, and the y-coordinate of S from R, because they are opposite corners.

🎯 Exam Tip: Remember the properties of geometric shapes like rectangles (e.g., opposite sides are parallel, vertices share coordinates) to find missing points efficiently without complex calculations.

 

Question 6. A line passes through (6, 0) and (0, 6) and another line passes through (-3, 0) and (0, -3). What are the points to be joined to get a trapezium?
Answer:
A trapezium is a four-sided shape (quadrilateral) with at least one pair of parallel sides. The other two sides are not parallel. This definition is key to forming the shape.
Let's consider the two lines formed by the given points:
Line 1: Joining (6, 0) and (0, 6).
Line 2: Joining (-3, 0) and (0, -3).
To check if these lines are parallel, we can look at their slopes. The slope \( m \) of a line passing through \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
Slope of Line 1: \( m_1 = \frac{6 - 0}{0 - 6} = \frac{6}{-6} = -1 \).
Slope of Line 2: \( m_2 = \frac{-3 - 0}{0 - (-3)} = \frac{-3}{3} = -1 \).
Since \( m_1 = m_2 \), the two lines are parallel.
The points defining these two parallel lines are (0, 6), (6, 0), (-3, 0), and (0, -3). To form a trapezium using these, we must connect the endpoints of these parallel segments in a way that creates two non-parallel sides.
The points to be joined to form a trapezium are (0, 6), (6, 0), (0, -3), and (-3, 0). You can join (0, 6) to (6, 0) and (-3, 0) to (0, -3) to get two parallel lines. Then connect (0, 6) to (0, -3) and (6, 0) to (-3, 0) to form the non-parallel sides. Or connect (0, 6) to (-3,0) and (6,0) to (0,-3).

In simple words: We are given four points and asked how to connect them to make a trapezium, which is a shape with two parallel sides. We found that two sets of points make parallel lines, so we connect the other points to create the two non-parallel sides.

🎯 Exam Tip: To identify parallel lines, calculate their slopes. If the slopes are equal, the lines are parallel. This is a fundamental concept in coordinate geometry.

 

Question 7. Find the point of intersection of the line joining points (- 3, 7) (2, – 4) and (4, 6) (- 5, – 7). Also find the point of intersection of these lines and also their intersection with the axis.
Answer:
First, we find the equation of Line 1, which joins \( (-3, 7) \) and \( (2, -4) \). We use the two-point formula \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \):
\( \frac{y - 7}{-4 - 7} = \frac{x - (-3)}{2 - (-3)} \)
\( \frac{y - 7}{-11} = \frac{x + 3}{5} \)
Now, cross-multiply:
\( 5(y - 7) = -11(x + 3) \)
\( 5y - 35 = -11x - 33 \)
Rearrange the terms to get the linear equation:
\( 11x + 5y = 35 - 33 \)
\( 11x + 5y = 2 \) (Equation 1)

Next, we find the equation of Line 2, which joins \( (4, 6) \) and \( (-5, -7) \). Using the same two-point formula:
\( \frac{y - 6}{-7 - 6} = \frac{x - 4}{-5 - 4} \)
\( \frac{y - 6}{-13} = \frac{x - 4}{-9} \)
Now, cross-multiply:
\( -9(y - 6) = -13(x - 4) \)
\( -9y + 54 = -13x + 52 \)
Rearrange the terms:
\( 13x - 9y = 52 - 54 \)
\( 13x - 9y = -2 \) (Equation 2)

To find the point of intersection of Line 1 and Line 2, we solve the system of equations:
1. \( 11x + 5y = 2 \)
2. \( 13x - 9y = -2 \)
Multiply Equation 1 by 9 and Equation 2 by 5 to eliminate y:
\( 9(11x + 5y) = 9(2) \implies 99x + 45y = 18 \) (Equation 3)
\( 5(13x - 9y) = 5(-2) \implies 65x - 45y = -10 \) (Equation 4)
Add Equation 3 and Equation 4:
\( (99x + 45y) + (65x - 45y) = 18 + (-10) \)
\( 164x = 8 \)
\( x = \frac{8}{164} \)
\( x = \frac{2}{41} \)
Substitute \( x = \frac{2}{41} \) into Equation 1:
\( 11(\frac{2}{41}) + 5y = 2 \)
\( \frac{22}{41} + 5y = 2 \)
\( 5y = 2 - \frac{22}{41} \)
\( 5y = \frac{82 - 22}{41} \)
\( 5y = \frac{60}{41} \)
\( y = \frac{60}{41 \times 5} \)
\( y = \frac{12}{41} \)
The point of intersection of the two lines is \( (\frac{2}{41}, \frac{12}{41}) \). This point satisfies both line equations.

Now, we find the points where each line intersects the x-axis and y-axis:
**For Line 1: \( 11x + 5y = 2 \)**
Intersection with x-axis (where \( y = 0 \)):
\( 11x + 5(0) = 2 \)
\( 11x = 2 \)
\( x = \frac{2}{11} \)
Point: \( (\frac{2}{11}, 0) \)
Intersection with y-axis (where \( x = 0 \)):
\( 11(0) + 5y = 2 \)
\( 5y = 2 \)
\( y = \frac{2}{5} \)
Point: \( (0, \frac{2}{5}) \)

**For Line 2: \( 13x - 9y = -2 \)**
Intersection with x-axis (where \( y = 0 \)):
\( 13x - 9(0) = -2 \)
\( 13x = -2 \)
\( x = -\frac{2}{13} \)
Point: \( (-\frac{2}{13}, 0) \)
Intersection with y-axis (where \( x = 0 \)):
\( 13(0) - 9y = -2 \)
\( -9y = -2 \)
\( y = \frac{2}{9} \)
Point: \( (0, \frac{2}{9}) \)
In simple words: We first find the mathematical rule for each of the two lines. Then we use these two rules together to find the one point where both lines cross each other. After that, for each line, we figure out where it crosses the horizontal (x) line and the vertical (y) line on the graph.

🎯 Exam Tip: To find the intersection of two lines, solve their equations simultaneously. To find axis intercepts, set the other coordinate to zero (e.g., set \( y=0 \) for x-intercept, \( x=0 \) for y-intercept).

 

Question 8. Draw the graph of the following equations: (i) \( x = -7 \) (ii) \( y = 6 \)
Answer:
(i) The equation \( x = -7 \) represents a vertical line passing through \( -7 \) on the x-axis. All points on this line have an x-coordinate of -7, regardless of their y-coordinate. Such lines are always parallel to the y-axis.
(ii) The equation \( y = 6 \) represents a horizontal line passing through \( 6 \) on the y-axis. All points on this line have a y-coordinate of 6, regardless of their x-coordinate. These lines are always parallel to the x-axis.

In simple words: For \( x = -7 \), draw a straight line that goes up and down, crossing the x-axis at the -7 mark. For \( y = 6 \), draw a straight line that goes left and right, crossing the y-axis at the 6 mark.

🎯 Exam Tip: Remember that equations like \( x = a \) always represent vertical lines, and equations like \( y = b \) always represent horizontal lines on a graph.

 

Question 9. Draw the graph of
(i) \( y = -3x \)
(ii) \( y = x - 4 \)
(iii) \( y = 2x + 5 \)
Answer:
To draw the graph of each linear equation, we need to find at least two points that satisfy the equation. We can do this by substituting simple values for x and calculating the corresponding y-values. Plotting these points and connecting them with a straight line will give the graph of the equation.

**(i) For \( y = -3x \):**
- If \( x = 0 \), then \( y = -3 \times 0 = 0 \). Point: (0, 0)
- If \( x = 1 \), then \( y = -3 \times 1 = -3 \). Point: (1, -3)
- If \( x = -1 \), then \( y = -3 \times (-1) = 3 \). Point: (-1, 3)

**(ii) For \( y = x - 4 \):**
- If \( x = 0 \), then \( y = 0 - 4 = -4 \). Point: (0, -4)
- If \( x = 4 \), then \( y = 4 - 4 = 0 \). Point: (4, 0)
- If \( x = 2 \), then \( y = 2 - 4 = -2 \). Point: (2, -2)

**(iii) For \( y = 2x + 5 \):**
- If \( x = -1 \), then \( y = 2(-1) + 5 = -2 + 5 = 3 \). Point: (-1, 3)
- If \( x = -2 \), then \( y = 2(-2) + 5 = -4 + 5 = 1 \). Point: (-2, 1)
- If \( x = 0 \), then \( y = 2(0) + 5 = 5 \). Point: (0, 5)

In simple words: For each equation, choose a few easy numbers for 'x' and then figure out what 'y' would be. Then, draw these points on a graph and connect them with a straight line. This line is the graph of the equation.

🎯 Exam Tip: When graphing linear equations, always find at least two points (intercepts are often easiest) and check a third point to ensure accuracy.

 

Question 10. Find the values
(a) \( y = x + 3 \)
(b) \( 2x + y - 6 = 0 \)
(c) \( y = 3x + 1 \)
Answer:
**(a) For \( y = x + 3 \)**
We are finding the missing values in the table for the equation \( y = x + 3 \). This equation shows a direct relationship between x and y, where y is always 3 more than x.
- If \( x = 0 \), then \( y = 0 + 3 = 3 \).
- If \( y = 0 \), then \( 0 = x + 3 \implies x = -3 \).
- If \( x = -2 \), then \( y = -2 + 3 = 1 \).
- If \( y = -3 \), then \( -3 = x + 3 \implies x = -6 \).

**(b) For \( 2x + y - 6 = 0 \)**
We are finding the missing values in the table for the equation \( 2x + y - 6 = 0 \), which can be rewritten as \( y = 6 - 2x \).
- If \( x = 0 \), then \( 2(0) + y - 6 = 0 \implies y = 6 \).
- If \( y = 0 \), then \( 2x + 0 - 6 = 0 \implies 2x = 6 \implies x = 3 \).
- If \( x = -1 \), then \( 2(-1) + y - 6 = 0 \implies -2 + y - 6 = 0 \implies y = 8 \).
- If \( y = -2 \), then \( 2x + (-2) - 6 = 0 \implies 2x - 8 = 0 \implies 2x = 8 \implies x = 4 \).

**(c) For \( y = 3x + 1 \)**
We are finding the missing values in the table for the equation \( y = 3x + 1 \). This equation shows that y is one more than three times x.
- If \( x = -1 \), then \( y = 3(-1) + 1 = -3 + 1 = -2 \).
- If \( x = 0 \), then \( y = 3(0) + 1 = 1 \).
- If \( x = 1 \), then \( y = 3(1) + 1 = 3 + 1 = 4 \).
- If \( x = 2 \), then \( y = 3(2) + 1 = 6 + 1 = 7 \).

In simple words: For each rule (equation), we are given a table with some missing 'x' or 'y' values. We need to use the rule to find the missing number for each pair so that the numbers correctly follow the equation.

🎯 Exam Tip: To find missing values in a table for a given linear equation, simply substitute the known x-value to find y, or the known y-value to find x, and solve the resulting simple equation.

 

Question 1. Fill in the blanks:
(i) \( y = px \) where \( p \in Z \) always passes through the ______.
Answer: Origin (0,0). A line in the form \( y = px \) represents a direct variation. When \( x = 0 \), then \( y = p \times 0 = 0 \). Therefore, the point \( (0,0) \) is always a solution for such an equation.
(ii) The intersecting point of the line \( x = 4 \) and \( y = -4 \) is ______.
Answer: \( (4, -4) \). The equation \( x = 4 \) describes a vertical line where the x-coordinate is always 4. The equation \( y = -4 \) describes a horizontal line where the y-coordinate is always -4. The point where these two lines meet must satisfy both conditions simultaneously.
(iii) Scale for the given graph, On the x-axis 1 cm = ______ units y-axis 1 cm = ______ units
Answer: For the x-axis, 1 cm = 3 units, and for the y-axis, 1 cm = 25 units. This specific scale helps in representing larger numerical values on a standard graph sheet in a compact and readable manner.

In simple words: This question checks your understanding of basic graph concepts. It involves recognizing that direct variation equations always pass through the starting point (origin), finding the intersection of simple vertical and horizontal lines, and correctly stating the measurement scale used on a graph.

🎯 Exam Tip: For fill-in-the-blank questions related to graphs and coordinate geometry, always refer to fundamental definitions and properties. A line \( y=mx \) always passes through the origin \( (0,0) \). Lines \( x=a \) and \( y=b \) intersect at \( (a,b) \). Always clearly state the scale when interpreting graphs.

 

Question 2.
(i) The points \( (1,1) \), \( (2,2) \), \( (3,3) \) lie on a same straight line.
Answer: True. All these points share a property where the y-coordinate is equal to the x-coordinate. This means they all satisfy the equation \( y = x \), which is a linear equation representing a straight line passing through the origin. Since all three points satisfy the same linear equation, they must lie on the same straight line.
(ii) \( y = -9x \) not passes through the origin.
Answer: False. To check if a line passes through the origin, we substitute \( x = 0 \) and \( y = 0 \) into the equation. For \( y = -9x \), if we substitute \( x = 0 \), we get \( y = -9 \times 0 = 0 \). This means that the point \( (0,0) \), which is the origin, lies on the line. Therefore, the statement that it does not pass through the origin is incorrect.
In simple words: This question tests your knowledge of straight lines. Points where \( y \) equals \( x \), like \( (1,1) \) or \( (2,2) \), are always on the same straight line. Also, any equation like \( y = \text{number} \times x \) will always go through the very center of the graph, which is called the origin \( (0,0) \).

🎯 Exam Tip: To verify if points lie on a line, substitute their coordinates into the line's equation. If the equation holds true, the points are collinear. For a line to pass through the origin, setting \( x=0 \) must result in \( y=0 \).

 

Question 3. Will a line pass through \( (2, 2) \) if it intersects the axes at \( (2, 0) \) and \( (0, 2) \)?
Answer: No, the line will not pass through \( (2, 2) \). We can find the equation of the line using the intercept form, which is \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) is the x-intercept and \( b \) is the y-intercept. Given the intercepts are \( (2, 0) \) and \( (0, 2) \), we have \( a = 2 \) and \( b = 2 \). Substituting these values into the intercept form, the equation of the line becomes \( \frac{x}{2} + \frac{y}{2} = 1 \). Multiplying the entire equation by 2, we get \( x + y = 2 \). Now, to check if the point \( (2, 2) \) lies on this line, we substitute its coordinates into the line's equation: \( 2 + 2 = 4 \). Since \( 4 \ne 2 \), the point \( (2, 2) \) does not satisfy the equation of the line, meaning it does not lie on the line. Understanding the intercept form simplifies finding the line's equation quickly.
In simple words: No, the line does not pass through \( (2,2) \). If a line crosses the x-axis at \( (2,0) \) and the y-axis at \( (0,2) \), its rule (equation) is \( x+y=2 \). When you try to put \( x=2 \) and \( y=2 \) into this rule, you get \( 2+2=4 \), which is not equal to \( 2 \). So, the point \( (2,2) \) is not on this line.

🎯 Exam Tip: When given the x and y intercepts, the intercept form of a line \( \frac{x}{a} + \frac{y}{b} = 1 \) is very efficient for finding the equation. To check if any point lies on a line, always substitute its coordinates into the line's equation; if it satisfies the equation, it's on the line.

 

Question 4. A line passing through \( (4, -2) \) and intersects the Y-axis at \( (0, 2) \). Find a point on the line in the second quadrant.
Answer: First, we need to determine the equation of the line using the two given points, \( (x_1, y_1) = (4, -2) \) and \( (x_2, y_2) = (0, 2) \). We will use the two-point form of a linear equation: \( \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \).
Substituting the coordinates:
\( \frac{y - (-2)}{x - 4} = \frac{2 - (-2)}{0 - 4} \)
\( \frac{y + 2}{x - 4} = \frac{4}{-4} \)
\( \frac{y + 2}{x - 4} = -1 \)
Now, we cross-multiply to simplify the equation:
\( y + 2 = -1 \times (x - 4) \)
\( y + 2 = -x + 4 \)
Rearranging the terms to find the standard form of the line's equation:
\( x + y = 4 - 2 \)
\( x + y = 2 \). This is the equation of the line. All points on this line must satisfy this equation.
The second quadrant is defined by points where the x-coordinate is negative \( (x < 0) \) and the y-coordinate is positive \( (y > 0) \). To find such a point, we can choose any negative value for \( x \) and substitute it into the line's equation. Let's choose \( x = -2 \).
Substitute \( x = -2 \) into \( x + y = 2 \):
\( -2 + y = 2 \)
\( y = 2 + 2 \)
\( y = 4 \).
So, a point on the line that lies in the second quadrant is \( (-2, 4) \). This point meets both conditions: it is on the line and in the correct quadrant.

In simple words: First, we find the rule (equation) of the line using the two points given. Then, we look for a point on this line that is in the second quadrant, which means its x-value must be negative and its y-value positive. By choosing \( x=-2 \), we found \( y=4 \), so \( (-2,4) \) is the point we need.

🎯 Exam Tip: To find the equation of a line given two points, use the two-point formula. Remember the sign conventions for x and y coordinates in each quadrant: Q1 (+,+), Q2 (-,+), Q3 (-,-), Q4 (+,-). This helps in identifying points in specific regions.

 

Question 5. If the points \( P(5, 3) \), \( Q(-3, 3) \), \( R(-3, -4) \) and \( S \) form a rectangle, then find the coordinate of \( S \).
Answer: To find the coordinates of point \( S \), we can use the properties of a rectangle. In a rectangle, opposite sides are parallel and equal in length, and its adjacent sides are perpendicular.
Given points are: \( P(5, 3) \), \( Q(-3, 3) \), \( R(-3, -4) \).
Let's analyze the given points:
Points \( P(5, 3) \) and \( Q(-3, 3) \) have the same y-coordinate \( (y=3) \). This indicates that the side \( PQ \) is a horizontal line.
Points \( Q(-3, 3) \) and \( R(-3, -4) \) have the same x-coordinate \( (x=-3) \). This indicates that the side \( QR \) is a vertical line.
Since \( PQ \) is horizontal and \( QR \) is vertical, they are perpendicular, which is correct for adjacent sides of a rectangle.
For the rectangle \( PQRS \), the side \( RS \) must be parallel to \( PQ \), and the side \( PS \) must be parallel to \( QR \).
Since \( RS \) is parallel to \( PQ \), \( RS \) must also be a horizontal line. As \( R \) has a y-coordinate of \( -4 \), \( S \) must also have a y-coordinate of \( -4 \). So, \( y_S = -4 \).
Since \( PS \) is parallel to \( QR \), \( PS \) must also be a vertical line. As \( P \) has an x-coordinate of \( 5 \), \( S \) must also have an x-coordinate of \( 5 \). So, \( x_S = 5 \).
Therefore, the coordinate of point \( S \) is \( (5, -4) \). This completes the rectangle by forming the missing vertex.

In simple words: For a rectangle, opposite sides are always parallel. Since \( P \) and \( Q \) have the same 'y' coordinate (making \( PQ \) a flat line), the missing point \( S \) must have the same 'x' coordinate as \( P \) and the same 'y' coordinate as \( R \). So, \( S \) is at \( (5, -4) \).

🎯 Exam Tip: When finding missing coordinates of a rectangle or parallelogram, use the property that opposite sides are parallel. This means they either share the same x-coordinate (for vertical sides) or the same y-coordinate (for horizontal sides).

 

Question 6. Given the points \( (0, 6) \), \( (6, 0) \), \( (-3, 0) \), and \( (0, -3) \). What are the points to be joined to get a trapezium?
Answer: A trapezium (also known as a trapezoid) is a quadrilateral with exactly one pair of parallel opposite sides. To identify which points form a trapezium, we need to check the slopes of the line segments formed by connecting these points. Let's label the points as \( A(0, 6) \), \( B(6, 0) \), \( C(-3, 0) \), and \( D(0, -3) \).
The formula for the slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
Slope of \( AB = \frac{0 - 6}{6 - 0} = \frac{-6}{6} = -1 \)
Slope of \( BC = \frac{0 - 0}{-3 - 6} = \frac{0}{-9} = 0 \)
Slope of \( CD = \frac{-3 - 0}{0 - (-3)} = \frac{-3}{3} = -1 \)
Slope of \( DA = \frac{6 - (-3)}{0 - 0} = \frac{9}{0} \) (Undefined, meaning vertical line)
From the calculations, we can see that the slope of \( AB \) is \( -1 \) and the slope of \( CD \) is also \( -1 \). Since these slopes are equal, line segments \( AB \) and \( CD \) are parallel to each other. The other two sides, \( BC \) and \( DA \), are not parallel (one is horizontal, one is vertical).
Therefore, to form a trapezium, we should join the points in an order that makes \( AB \) and \( CD \) the parallel sides. A possible order is \( A(0, 6) \), \( B(6, 0) \), \( D(0, -3) \), and \( C(-3, 0) \).

In simple words: To make a trapezium, we need two opposite sides to be parallel. We checked the slopes (steepness) of all possible lines between the given points. We found that the line from \( (0,6) \) to \( (6,0) \) and the line from \( (-3,0) \) to \( (0,-3) \) both have a slope of \( -1 \), meaning they are parallel. So, joining points \( (0,6) \), \( (6,0) \), \( (0,-3) \), and \( (-3,0) \) in order forms a trapezium.

🎯 Exam Tip: The key to identifying a trapezium is to calculate the slopes of all four potential sides. Parallel lines always have equal slopes. Remember that a vertical line has an undefined slope, and a horizontal line has a slope of zero.

 

Question 7. Find the point of intersection of the line joining points \( (-3, 7) \) and \( (2, -4) \) with the line joining points \( (4, 6) \) and \( (-5, -7) \). Also, find the point of intersection of each of these lines with the x and y axes.
Answer: This problem requires multiple steps: first, finding the equation for each of the two lines; second, solving these two equations simultaneously to find their intersection point; and third, determining where each line crosses the x and y axes. We will use the two-point formula for a line, which is \( \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \).

Part 1: Equation of Line 1 (joining points \( (-3, 7) \) and \( (2, -4) \))
Let \( (x_1, y_1) = (-3, 7) \) and \( (x_2, y_2) = (2, -4) \).
\( \frac{y - 7}{x - (-3)} = \frac{-4 - 7}{2 - (-3)} \)
\( \frac{y - 7}{x + 3} = \frac{-11}{5} \)
Cross-multiplying to eliminate the denominators:
\( 5(y - 7) = -11(x + 3) \)
\( 5y - 35 = -11x - 33 \)
Rearranging the terms to form a linear equation in standard form:
\( 11x + 5y = 35 - 33 \)
\( 11x + 5y = 2 \) (Equation 1)

Part 2: Equation of Line 2 (joining points \( (4, 6) \) and \( (-5, -7) \))
Let \( (x_1, y_1) = (4, 6) \) and \( (x_2, y_2) = (-5, -7) \).
\( \frac{y - 6}{x - 4} = \frac{-7 - 6}{-5 - 4} \)
\( \frac{y - 6}{x - 4} = \frac{-13}{-9} \)
\( \frac{y - 6}{x - 4} = \frac{13}{9} \)
Cross-multiplying:
\( 9(y - 6) = 13(x - 4) \)
\( 9y - 54 = 13x - 52 \)
Rearranging terms to standard form:
\( 9y - 13x = 54 - 52 \)
\( -13x + 9y = 2 \) (Equation 2)

Part 3: Point of Intersection of Line 1 and Line 2
We solve the system of linear equations using the elimination method:
1) \( 11x + 5y = 2 \)
2) \( -13x + 9y = 2 \)
To eliminate \( x \), multiply Equation 1 by 13 and Equation 2 by 11:
\( 13 \times (11x + 5y) = 13 \times 2 \implies 143x + 65y = 26 \) (Equation 3)
\( 11 \times (-13x + 9y) = 11 \times 2 \implies -143x + 99y = 22 \) (Equation 4)
Add Equation 3 and Equation 4:
\( (143x + 65y) + (-143x + 99y) = 26 + 22 \)
\( 164y = 48 \)
\( y = \frac{48}{164} = \frac{12}{41} \)
Substitute the value of \( y \) into Equation 1 to find \( x \):
\( 11x + 5 \left( \frac{12}{41} \right) = 2 \)
\( 11x + \frac{60}{41} = 2 \)
\( 11x = 2 - \frac{60}{41} \)
\( 11x = \frac{82 - 60}{41} \)
\( 11x = \frac{22}{41} \)
\( x = \frac{22}{41 \times 11} \)
\( x = \frac{2}{41} \)
The point of intersection of the two lines is \( \left( \frac{2}{41}, \frac{12}{41} \right) \). This is the unique point that lies on both lines.

Part 4: Intersection of Line 1 with the x and y axes
Equation of Line 1: \( 11x + 5y = 2 \)
To find the x-intercept (where \( y=0 \)):
\( 11x + 5(0) = 2 \implies 11x = 2 \implies x = \frac{2}{11} \)
The x-intercept for Line 1 is \( \left( \frac{2}{11}, 0 \right) \).
To find the y-intercept (where \( x=0 \)):
\( 11(0) + 5y = 2 \implies 5y = 2 \implies y = \frac{2}{5} \)
The y-intercept for Line 1 is \( \left( 0, \frac{2}{5} \right) \).

Part 5: Intersection of Line 2 with the x and y axes
Equation of Line 2: \( -13x + 9y = 2 \)
To find the x-intercept (where \( y=0 \)):
\( -13x + 9(0) = 2 \implies -13x = 2 \implies x = \frac{-2}{13} \)
The x-intercept for Line 2 is \( \left( \frac{-2}{13}, 0 \right) \).
To find the y-intercept (where \( x=0 \)):
\( -13(0) + 9y = 2 \implies 9y = 2 \implies y = \frac{2}{9} \)
The y-intercept for Line 2 is \( \left( 0, \frac{2}{9} \right) \).
In simple words: We first find the mathematical rule (equation) for each of the two lines using the points they pass through. Then, we solve these two rules together to find the one point where the lines cross. After that, for each line, we find where it cuts the x-axis (by setting y to zero) and where it cuts the y-axis (by setting x to zero). This process gives us all the required intersection points.

🎯 Exam Tip: When dealing with multiple line intersection problems, organize your steps clearly. Always start by finding the equation of each line. For intersections, simultaneous equations are key. For axis intercepts, remember that \( y=0 \) on the x-axis and \( x=0 \) on the y-axis.

 

Question 8. Draw the graph of the following equations:
(i) \( x = -7 \)
(ii) \( y = 6 \)
Answer: To draw the graph of these equations, it's important to remember their characteristics. An equation of the form \( x = \text{constant} \) represents a vertical line, and an equation of the form \( y = \text{constant} \) represents a horizontal line. We plot these lines on a coordinate plane.
(i) For \( x = -7 \): This is a vertical line that passes through every point where the x-coordinate is \( -7 \), regardless of the y-coordinate. It is parallel to the y-axis.
(ii) For \( y = 6 \): This is a horizontal line that passes through every point where the y-coordinate is \( 6 \), regardless of the x-coordinate. It is parallel to the x-axis. These types of lines form the basic building blocks of coordinate geometry graphing.

In simple words: To draw these graphs, remember that \( x = -7 \) means a straight line going up and down (vertical) through the number -7 on the x-axis. And \( y = 6 \) means a straight line going sideways (horizontal) through the number 6 on the y-axis.

🎯 Exam Tip: Remember that equations of the form \( x=c \) always produce vertical lines, and equations of the form \( y=c \) always produce horizontal lines. These lines are parallel to the y-axis and x-axis respectively.

 

Question 9. Draw the graph of the following equations:
(i) \( y = -3x \)
(ii) \( y = x - 4 \)
(iii) \( y = 2x + 5 \)
Answer: To draw the graph for each linear equation, we need to identify at least two points that lie on each line. This is done by substituting different values for \( x \) and calculating the corresponding \( y \) values. Once we have these points, we plot them on a coordinate plane and draw a straight line through them, as linear equations always represent straight lines.

(i) For \( y = -3x \):
If \( x = 0 \), then \( y = -3 \times 0 = 0 \). So, one point is \( (0, 0) \).
If \( x = 1 \), then \( y = -3 \times 1 = -3 \). So, another point is \( (1, -3) \).

(ii) For \( y = x - 4 \):
If \( x = 0 \), then \( y = 0 - 4 = -4 \). So, one point is \( (0, -4) \).
If \( x = 4 \), then \( y = 4 - 4 = 0 \). So, another point is \( (4, 0) \).

(iii) For \( y = 2x + 5 \):
If \( x = -1 \), then \( y = 2(-1) + 5 = -2 + 5 = 3 \). So, one point is \( (-1, 3) \).
If \( x = -2 \), then \( y = 2(-2) + 5 = -4 + 5 = 1 \). So, another point is \( (-2, 1) \).
We plot these calculated points on the graph and connect them with straight lines to represent each equation.

In simple words: To draw these graphs, you need to find two points for each equation that make the equation true. For example, for \( y = -3x \), if \( x=0 \), \( y=0 \), and if \( x=1 \), \( y=-3 \). Plot these points, then draw a straight line through them. Do this for all three equations.

🎯 Exam Tip: To graph linear equations efficiently, find the y-intercept (by setting \( x=0 \)) and one other point. If the equation passes through the origin (like \( y=-3x \)), choose two distinct non-origin points for better accuracy in drawing the line.

 

Question 10. Find the missing values in the tables for each equation:
(a) \( y = x + 3 \)

Answer: For the equation \( y = x + 3 \), we substitute the given \( x \) or \( y \) values to find the missing ones:
1. If \( x = 0 \), then \( y = 0 + 3 = 3 \).
2. If \( y = 0 \), then \( 0 = x + 3 \implies x = -3 \).
3. If \( x = -2 \), then \( y = -2 + 3 = 1 \).
4. If \( y = -3 \), then \( -3 = x + 3 \implies x = -6 \).
The completed table is:

(b) \( 2x + y - 6 = 0 \)

Answer: For the equation \( 2x + y - 6 = 0 \) (which can also be written as \( y = -2x + 6 \)), we find the missing values:
1. If \( x = 0 \), then \( 2(0) + y - 6 = 0 \implies y - 6 = 0 \implies y = 6 \).
2. If \( y = 0 \), then \( 2x + 0 - 6 = 0 \implies 2x = 6 \implies x = 3 \).
3. If \( x = -1 \), then \( 2(-1) + y - 6 = 0 \implies -2 + y - 6 = 0 \implies y - 8 = 0 \implies y = 8 \).
4. If \( y = -2 \), then \( 2x + (-2) - 6 = 0 \implies 2x - 8 = 0 \implies 2x = 8 \implies x = 4 \).
The completed table is:

(c) \( y = 3x + 1 \)

Answer: For the equation \( y = 3x + 1 \), we calculate the corresponding \( y \) values for the given \( x \) values:
1. If \( x = -1 \), then \( y = 3(-1) + 1 = -3 + 1 = -2 \).
2. If \( x = 0 \), then \( y = 3(0) + 1 = 0 + 1 = 1 \).
3. If \( x = 1 \), then \( y = 3(1) + 1 = 3 + 1 = 4 \).
4. If \( x = 2 \), then \( y = 3(2) + 1 = 6 + 1 = 7 \).
The completed table is:

In simple words: To fill in these tables, you use the given rule (equation) for each part. If you have an \( x \) value, put it into the equation to find its matching \( y \) value. If you have a \( y \) value, put it into the equation to find its matching \( x \) value. Every pair of \( x \) and \( y \) numbers must make the equation true.

🎯 Exam Tip: When completing tables for linear equations, substitute values carefully into the equation to avoid calculation errors. Writing the equation in the form \( y = mx + c \) can often simplify the process of finding \( y \) for given \( x \) values.