OP Malhotra Class 9 Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Exercise 20 (C)

S Chand Class 9 ICSE Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Ex 20(C)

Question 1. 3y – 2x = 7, 5x + 3y = -7
Answer: For the first equation, \( 3y - 2x = 7 \), we can rearrange it to find \( y \):
\( 3y = 7 + 2x \)
\( \implies y = \frac{7+2x}{3} \)
Now, we choose different values for \( x \) and calculate the corresponding values for \( y \):

We plot these points (1, 3), (4, 5), and (-2, 1) on a graph and draw a line through them. This line represents the first equation.
For the second equation, \( 5x + 3y = -7 \), we rearrange it to find \( x \):
\( 5x = -7 - 3y \)
\( \implies x = \frac{-7-3y}{5} \)
\( \implies x = -\left(\frac{7+3y}{5}\right) \)
Next, we choose different values for \( y \) and calculate the corresponding values for \( x \):

We then plot these points (-2, 1), (-5, 6), and (1, -4) on the same graph and draw another line. When we look at the graph, we can see that these two lines cross each other at the point (-2, 1). This point is the solution to both equations. Drawing graphs helps visualize the solution of simultaneous equations.
Therefore, the solution is \( x = -2 \) and \( y = 1 \).
In simple words: We find points for each line by choosing values, then plot these points on a graph. Where the two lines cross, that is the answer for both equations.

🎯 Exam Tip: Always make sure your graph paper has clearly marked axes and a consistent scale. Plot at least three points for each line to ensure accuracy when drawing straight lines, as two points are sufficient but can lead to errors if one point is miscalculated or plotted incorrectly.

Question 2. 2x + 3y = 13, 5x - 2y = 4
Answer: For the first equation, \( 2x + 3y = 13 \), we rearrange it to find \( x \):
\( 2x = 13 - 3y \)
\( \implies x = \frac{13-3y}{2} \)
We choose different values for \( y \) and calculate the corresponding values for \( x \):

We plot these points (5, 1), (2, 3), and (-1, 5) on a graph and draw a line. This line represents the first equation.
For the second equation, \( 5x - 2y = 4 \), we rearrange it to find \( x \):
\( 5x = 4 + 2y \)
\( \implies x = \frac{4+2y}{5} \)
Now, we choose different values for \( y \) and calculate the corresponding values for \( x \):

We plot these points (2, 3), (0, -2), and (-2, -7) on the same graph and draw the second line. The two lines intersect each other at the point (2, 3). This point is the solution for the system of equations. Graphing helps to visually confirm the intersection, which is the common solution.
Therefore, the solution is \( x = 2 \) and \( y = 3 \).
In simple words: First, rewrite each equation to easily find points. Then, draw both lines on the same graph. The spot where they cross is the answer.

🎯 Exam Tip: When plotting points, always label them on the graph. This helps in cross-checking and ensures clarity, especially when multiple lines are involved. Using different colors for each line can also make your graph easier to understand.

Question 3. 5x + y = -3, 2x = 3y – 8
Answer: For the first equation, \( 5x + y = -3 \), we rearrange it to find \( y \):
\( y = -3 - 5x \)
\( \implies y = -(3 + 5x) \)
Now, we choose different values for \( x \) and calculate the corresponding values for \( y \):

We plot these points (0, -3), (-1, 2), and (-2, 7) on a graph and draw the first line.
For the second equation, \( 2x = 3y - 8 \), we rearrange it to find \( x \):
\( x = \frac{3y-8}{2} \)
Next, we choose different values for \( y \) and calculate the corresponding values for \( x \):

We plot these points (-4, 0), (-1, 2), and (2, 4) on the same graph and draw the second line. The two lines intersect each other at the point (-1, 2). This intersection point gives us the solution for both equations. By finding the \(x\) and \(y\) values where the lines cross, we solve the system.
Therefore, the solution is \( x = -1 \) and \( y = 2 \).
In simple words: Get a few points for each line by plugging in numbers. Draw both lines on a graph. The spot where they cross is your answer.

🎯 Exam Tip: Always check your plotted points by substituting them back into the original equations to ensure they satisfy the equation. This helps prevent plotting errors.

Question 4. Find graphically the vertices of the triangle whose sides have the equations 2y – x = 8, 5y – x = 14 and y – 2x = 1 respectively. Take 1 cm = 1 unit on both the axes.
Answer: We have three equations that represent the sides of a triangle:
1. \( 2y - x = 8 \)
2. \( 5y - x = 14 \)
3. \( y - 2x = 1 \)

Let's find points for each equation:

For equation 1: \( 2y - x = 8 \)
\( 2y = 8 + x \)
\( \implies y = \frac{8+x}{2} \)
We choose different values for \( x \) and calculate the corresponding values for \( y \):

We plot the points (0, 4), (-2, 3), and (-4, 2) and draw the first line.

For equation 2: \( 5y - x = 14 \)
\( x = 5y - 14 \)
We choose different values for \( y \) and calculate the corresponding values for \( x \):

We plot the points (-4, 2), (1, 3), and (6, 4) and draw the second line.

For equation 3: \( y - 2x = 1 \)
\( y = 1 + 2x \)
We choose different values for \( x \) and calculate the corresponding values for \( y \):

We plot the points (0, 1), (1, 3), and (2, 5) and draw the third line.
By drawing these three lines on the same graph, we can see where they cross each other. The points where any two lines intersect are the vertices of the triangle. These intersection points are (-4, 2), (2, 5), and (1, 3). Understanding how to set up the coordinate system is important for accurate graphing.
Therefore, the vertices of the triangle are \( (-4, 2) \), \( (2, 5) \), and \( (1, 3) \).

🎯 Exam Tip: When finding vertices of a triangle formed by linear equations, identify the intersection points of each pair of lines. There should be exactly three unique intersection points to form a triangle.

Question 5. Using a scale of 1 cm to 1 unit for both the axes, draw the graphs of the following equations :
6y = 5x + 10, y = 5x – 15 From the graphs find :
(i) the coordinates of the point where the two lines intersect ;
(ii) the area of the triangle between the lines and the x-axis.
Answer: We have two equations:
1. \( 6y = 5x + 10 \)
2. \( y = 5x - 15 \)

Let's find points for each equation:

For equation 1: \( 6y = 5x + 10 \)
\( \implies y = \frac{5x+10}{6} \)
We choose different values for \( x \) and calculate the corresponding values for \( y \):

We plot these points (4, 5), (10, 10), and (-2, 0) and draw the first line.

For equation 2: \( y = 5x - 15 \)
We choose different values for \( x \) and calculate the corresponding values for \( y \):

We plot these points (2, -5), (3, 0), and (4, 5) and draw the second line.

When we plot both lines on the graph:

(i) We can see that the two lines intersect each other at the point \( (4, 5) \). This point is the solution where both equations are true.
(ii) The x-axis and the two lines form a triangle. The vertices of this triangle are the x-intercepts of the lines and their intersection point. The x-intercept of the first line is \( (-2, 0) \) (from its table). The x-intercept of the second line is \( (3, 0) \) (from its table). The intersection of the two lines is \( (4, 5) \).
Let's call the vertices A \( (-2, 0) \), C \( (3, 0) \), and B \( (4, 5) \).
The base of the triangle AC lies on the x-axis, and its length is the distance between \( (-2, 0) \) and \( (3, 0) \), which is \( |3 - (-2)| = 5 \) units.
The height of the triangle is the perpendicular distance from vertex B \( (4, 5) \) to the x-axis, which is the absolute value of the y-coordinate of B, so height \( = 5 \) units.
The area of a triangle is calculated as \( \frac{1}{2} \times \text{base} \times \text{height} \).
Area \( = \frac{1}{2} \times 5 \times 5 \)
\( = \frac{25}{2} \)
\( = 12.5 \) sq. units.
In simple words: Draw both lines using the points you found. The spot where they cross is the intersection. For the triangle, find where each line crosses the 'x' axis and use those points, plus the intersection, to make a triangle. Then find its area using the base and height.

🎯 Exam Tip: When calculating the area of a triangle on a graph, identify the base and height clearly. The base is often easiest to measure along an axis, and the height is the perpendicular distance from the opposite vertex to that base.

Question 6. The graph of a linear equation in x and y Solution: x + y + 3 = 0, and 3x – 2y + 4 = 0
Answer: We have two equations to solve graphically:
1. \( x + y + 3 = 0 \)
2. \( 3x - 2y + 4 = 0 \)

(i) For the first equation, \( x + y + 3 = 0 \), we rearrange it to find \( x \):
\( x = -y - 3 \)
\( \implies x = -(y + 3) \)
We choose different values for \( y \) and calculate the corresponding values for \( x \):

We plot these points (-3, 0), (-2, -1), and (-1, -2) and draw the first line.
For the second equation, \( 3x - 2y + 4 = 0 \), we rearrange it to find \( x \):
\( 3x = 2y - 4 \)
\( \implies x = \frac{2y-4}{3} \)
We choose different values for \( y \) and calculate the corresponding values for \( x \):

We plot these points (-2, -1), (0, 2), and (2, 5) and draw the second line.

(ii) We observe that these two lines intersect each other at the point P \( (-2, -1) \). This point is the unique solution where both equations are satisfied.
Therefore, the coordinates of P are \( (-2, -1) \).
(iii) Now, we need to join the origin (0,0) to point P \( (-2, -1) \) to form a line segment OP. We then measure the length of OP. The distance formula is \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
OP \( = \sqrt{(-2-0)^2 + (-1-0)^2} \)
\( = \sqrt{(-2)^2 + (-1)^2} \)
\( = \sqrt{4 + 1} \)
\( = \sqrt{5} \)
\( \approx 2.236 \)
So, OP \( \approx 2.2 \) units (approximately). It is important to measure accurately on the graph.
In simple words: First, find points for each equation and draw both lines on a graph. The point where the lines cross is the answer. Then, draw a line from the graph's center to this answer point and measure its length.

🎯 Exam Tip: When asked to measure a length from the origin, remember to use the distance formula if you don't have a physical ruler for the graph. If using a ruler, ensure the scale factor (e.g., 1 cm = 1 unit) is strictly followed.

Question 9. Solve graphically the simultaneous equations, x – 2y = 1; x + y = 4. Use 2 cm = 1 unit on both axes and plot only three points per line.
Answer: We have the two equations:
1. \( x - 2y = 1 \)
2. \( x + y = 4 \)

For equation 1: \( x - 2y = 1 \)
\( x = 1 + 2y \)
We choose different values for \( y \) and calculate the corresponding values for \( x \):

We plot these points (1, 0), (3, 1), and (5, 2) on the graph and draw the first line.

For equation 2: \( x + y = 4 \)
\( x = 4 - y \)
We choose different values for \( y \) and calculate the corresponding values for \( x \):

We plot these points (4, 0), (3, 1), and (2, 2) on the same graph and draw the second line. On the graph, we can see that these two lines intersect each other at the point (3, 1). This point represents the solution that satisfies both equations. Using a proper scale helps maintain accuracy in graphical solutions.
Therefore, the solution is \( x = 3 \) and \( y = 1 \).

🎯 Exam Tip: When a specific scale (e.g., 2 cm = 1 unit) is mentioned, ensure your graph paper's markings reflect this. Each unit on the axis should visually represent the specified physical distance to get full marks for graphical accuracy.

Question 10. Use a graph paper for this question. Draw graph of 2x – y – 1 =0 and 2x + y = 9 on the same axes. Use 2 cm = 1 unit on both axes and plot only 3 points for each line. Write down the coordinates of the point of intersection of the two lines.
Answer: We have the two equations:
1. \( 2x - y - 1 = 0 \)
2. \( 2x + y = 9 \)

For equation 1: \( 2x - y - 1 = 0 \)
\( y = 2x - 1 \)
We choose different values for \( x \) and calculate the corresponding values for \( y \):

We plot these points (1, 1), (2, 3), and (3, 5) on the graph and draw the first line.

For equation 2: \( 2x + y = 9 \)
\( y = 9 - 2x \)
We choose different values for \( x \) and calculate the corresponding values for \( y \):

We plot these points (2, 5), (3, 3), and (4, 1) on the same graph and draw the second line. Looking at the graph, we can see that these two lines intersect each other at the point (2.5, 4). This point is the solution for both equations. Using graphical methods helps us find the solution visually, which is especially useful for understanding the concept of simultaneous equations.
Therefore, the coordinates of the point of intersection are \( (2.5, 4) \).

🎯 Exam Tip: When the intersection point does not fall exactly on a grid line, carefully estimate its coordinates. It's often a good practice to verify this approximate graphical solution by solving the equations algebraically if time permits, to ensure precision.