OP Malhotra Class 9 Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Test

 

Question 1. If (1, 2) and (3, 8) are the extremes of a diagonal of a square, then the area of the square is
(a) 10 sq. units
(b) 20 sq. units
(c) 15 sq. units
(d) 8 sq. units
Answer: (b) 20 sq. units
In simple words: First, we find the length of the diagonal of the square using the distance formula between the two given points. After that, we use the formula for the area of a square when the diagonal is known, which is (diagonal squared) divided by 2. This helps us find the area quickly.

🎯 Exam Tip: Remember the two key formulas for this type of problem: the distance formula for the diagonal and the area formula of a square using its diagonal length.

 

Question 2. If the distance between two points (0, -5) and (x, 0) is 13 units, then x =
(a) 10
(b) ±10
(c) 12
(d) ± 12
Answer: (d) ± 12
Applying the distance formula between the points \( (0, -5) \) and \( (x, 0) \):
Distance \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( 13 = \sqrt{(x - 0)^2 + (0 - (-5))^2} \)
\( 13 = \sqrt{x^2 + (5)^2} \)
\( 13 = \sqrt{x^2 + 25} \)
Now, square both sides to remove the square root:
\( 13^2 = x^2 + 25 \)
\( 169 = x^2 + 25 \)
Subtract 25 from both sides:
\( x^2 = 169 - 25 \)
\( x^2 = 144 \)
Take the square root of both sides:
\( x = \pm \sqrt{144} \)
\( x = \pm 12 \)
In simple words: We use the distance formula to set up an equation with the unknown 'x'. Then, we square both sides to get rid of the square root and solve for 'x'. Remember that when you take a square root, the answer can be both positive and negative.

🎯 Exam Tip: Always remember that \( \sqrt{a^2} = \pm a \) when solving for a variable in a squared equation, as distance can be in either direction on a coordinate plane.

 

Question 3. If A (x, y) is equidistant from P (-3, 2) and Q (2, -3) then
(a) 2x = y
(b) x = -y
(c) x = 2y
(d) x = y
Answer: (d) x = y
If point A(x, y) is equidistant from P(-3, 2) and Q(2, -3), it means the distance AP is equal to the distance AQ.
So, \( AP^2 = AQ^2 \) (squaring both sides simplifies calculations by removing square roots).
Using the distance formula, \( (x_2 - x_1)^2 + (y_2 - y_1)^2 \):
For AP: \( (x - (-3))^2 + (y - 2)^2 = (x + 3)^2 + (y - 2)^2 \)
For AQ: \( (x - 2)^2 + (y - (-3))^2 = (x - 2)^2 + (y + 3)^2 \)
Setting them equal:
\( (x + 3)^2 + (y - 2)^2 = (x - 2)^2 + (y + 3)^2 \)
Expand the squares:
\( x^2 + 6x + 9 + y^2 - 4y + 4 = x^2 - 4x + 4 + y^2 + 6y + 9 \)
Subtract \( x^2 \) and \( y^2 \) from both sides:
\( 6x + 9 - 4y + 4 = -4x + 4 + 6y + 9 \)
Combine constant terms:
\( 6x - 4y + 13 = -4x + 6y + 13 \)
Subtract 13 from both sides:
\( 6x - 4y = -4x + 6y \)
Move all x terms to one side and y terms to the other:
\( 6x + 4x = 6y + 4y \)
\( 10x = 10y \)
Divide both sides by 10:
\( x = y \)
In simple words: When a point is the same distance from two other points, we can set up an equation where the distance from the first point to the third is equal to the distance from the second point to the third. We square both sides of the distance formula equation to remove the square roots, then solve the simplified equation to find the relationship between x and y.

🎯 Exam Tip: To avoid working with square roots, always square both sides of the distance equation (\( AP = AQ \implies AP^2 = AQ^2 \)) at the start of your calculation.

 

Question 4. The nearest point from the origin is
(a) (2,-3)
(b) (6,0)
(c) (-2,-1)
(d) (3,5)
Answer: (c) (-2,-1)
To find the nearest point from the origin (0, 0), we need to calculate the distance from (0, 0) to each given point and choose the smallest distance.
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
(a) Distance from (0,0) to (2, -3):
\( d = \sqrt{(2 - 0)^2 + (-3 - 0)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \)
(b) Distance from (0,0) to (6, 0):
\( d = \sqrt{(6 - 0)^2 + (0 - 0)^2} = \sqrt{6^2 + 0^2} = \sqrt{36} = 6 \)
(c) Distance from (0,0) to (-2, -1):
\( d = \sqrt{(-2 - 0)^2 + (-1 - 0)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \)
(d) Distance from (0,0) to (3, 5):
\( d = \sqrt{(3 - 0)^2 + (5 - 0)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \)
Comparing the distances: \( \sqrt{13} \approx 3.6 \), \( 6 \), \( \sqrt{5} \approx 2.2 \), \( \sqrt{34} \approx 5.8 \).
The smallest distance is \( \sqrt{5} \), which corresponds to point (-2, -1).
In simple words: To find which point is closest to the origin (the center point 0,0), we calculate the distance from the origin to each given point. The point with the smallest distance is the closest one.

🎯 Exam Tip: Instead of comparing square roots directly, you can compare the squared distances (\( d^2 \)) to find the smallest value, as it preserves the order of magnitude.

 

Question 5. The coordinates of the vertices of a side of square are (4, -3) and (-1, -5). Its area is
(a) \( 2 \sqrt{29} \) sq. units
(b) \( \frac{\sqrt{89}}{2} \) sq. units
(c) 10 sq. units
(d) 29 sq. units
Answer: (d) 29 sq. units
The given points (4, -3) and (-1, -5) are the coordinates of the ends of one side of the square. We can find the length of this side using the distance formula.
Length of a side \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Let \( (x_1, y_1) = (4, -3) \) and \( (x_2, y_2) = (-1, -5) \).
Side length \( = \sqrt{(-1 - 4)^2 + (-5 - (-3))^2} \)
\( = \sqrt{(-5)^2 + (-5 + 3)^2} \)
\( = \sqrt{(-5)^2 + (-2)^2} \)
\( = \sqrt{25 + 4} \)
\( = \sqrt{29} \) units.
The area of a square is given by the formula \( \text{Area} = (\text{Side length})^2 \).
Area \( = (\sqrt{29})^2 \)
Area \( = 29 \) sq. units.
In simple words: First, we use the distance formula to find how long one side of the square is, using the two given corner points. Then, to find the area of the square, we simply multiply the side length by itself.

🎯 Exam Tip: Always confirm if the given points represent a side or a diagonal, as this determines which geometric formula to apply next.

 

Question 6. Solve by graphing. 2x + y = -8, y = \( \frac{1}{3} \)x - 1
Answer:
We need to graph both linear equations on the same coordinate plane. First, we find a few points for each equation by choosing different values for x and calculating y.
For the first equation: \( 2x + y = -8 \implies y = -2x - 8 \)

Plot the points (-2, -4), (-4, 0), and (-5, 2) and draw a line through them.
For the second equation: \( y = \frac{1}{3}x - 1 \)

Plot the points (3, 0), (6, 1), and (-3, -2) and draw a line through them.
The solution to the system of equations is the point where the two lines intersect. From the graph, we can see that the lines intersect at the point (-3, -2).
Therefore, \( x = -3 \) and \( y = -2 \).

In simple words: To solve equations by drawing graphs, we first find at least two points for each line by choosing values for 'x' and calculating 'y'. Then, we plot these points and draw a straight line through them on the same graph paper. The place where the two lines cross each other is the answer, which tells us the 'x' and 'y' values that work for both equations.

🎯 Exam Tip: Always use a ruler and pencil for drawing lines and ensure your points are accurately plotted to find the correct intersection point on the graph.

 

Question 7. Draw the 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 per line. Read the coordinates of their point of intersection.
Answer:
To draw the graphs, we need to find at least three points for each linear equation.
For the first equation: \( 2x - y - 1 = 0 \implies y = 2x - 1 \)

Plot the points (1, 1), (2, 3), and (0, -1) and draw a line through them.
For the second equation: \( 2x + y = 9 \implies y = -2x + 9 \)

Plot the points (4, 1), (5, -1), and (6, -3) and draw a line through them.
From the graph, observe the point where the two lines cross. This point of intersection is \( \left(\frac{5}{2}, 4\right) \) or (2.5, 4).
Therefore, \( x = \frac{5}{2} \) and \( y = 4 \).

In simple words: First, rewrite each equation to find 'y' by itself. Then, pick three simple 'x' values for each equation and calculate the 'y' values to get three points. Plot these points on a graph and draw a line for each equation. The spot where the two lines cross gives you the 'x' and 'y' values that solve both equations at the same time.

🎯 Exam Tip: When graphing, ensure your scale (e.g., 2 cm = 1 unit) is consistent on both axes and clearly marked. Using fractional points for intersection is common, so mark them precisely.

 

Question 8. Choose the equation whose graph is given here
(a) y = x
(b) x + y = 0
(c) y = 2x
(d) 2 + 3y = 7x
Answer: (b) x + y = 0
Observe the points on the given graph. The graph passes through the origin (0, 0). It also contains points where the x-coordinate and y-coordinate are equal in magnitude but opposite in sign, such as (1, -1) and (-1, 1).
Let's check the given options:
(a) \( y = x \): This line passes through (0, 0), (1, 1), (2, 2). This does not match the graph.
(b) \( x + y = 0 \): This can be rewritten as \( y = -x \). This line passes through (0, 0), (1, -1), (-1, 1), (2, -2), (-2, 2). This exactly matches the points observed on the graph.
(c) \( y = 2x \): This line passes through (0, 0), (1, 2), (-1, -2). This does not match.
(d) \( 2 + 3y = 7x \): This can be rewritten as \( 3y = 7x - 2 \implies y = \frac{7}{3}x - \frac{2}{3} \). This line does not pass through the origin (0,0) and its points do not match the graph.
So, the equation \( x + y = 0 \) is the correct choice.
In simple words: Look at the graph and see which points it goes through. For example, it goes through (1, -1) and (-1, 1). Then, test each given equation by plugging in these points. The equation that works for all the points on the graph is the correct answer. This graph shows that the 'y' value is always the negative of the 'x' value.

🎯 Exam Tip: When identifying an equation from a graph, always check if the line passes through the origin (0,0) and test at least two other distinct points on the line against each option.

 

Question 9. The distance between the point (0, 0) and the intersecting point of the graphs of x = 3 and y = 4 is
(a) 4 units
(b) 3 units
(c) 2 units
(d) 5 units
Answer: (d) 5 units
The intersection point of the graphs \( x = 3 \) and \( y = 4 \) is simply (3, 4).
We need to find the distance between the origin (0, 0) and the point (3, 4).
Using the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Let \( (x_1, y_1) = (0, 0) \) and \( (x_2, y_2) = (3, 4) \).
Distance \( = \sqrt{(3 - 0)^2 + (4 - 0)^2} \)
\( = \sqrt{3^2 + 4^2} \)
\( = \sqrt{9 + 16} \)
\( = \sqrt{25} \)
\( = 5 \) units.
In simple words: First, find the point where the two lines x=3 and y=4 cross. This point is simply (3,4). Then, use the distance formula to measure how far this point is from the center point (0,0).

🎯 Exam Tip: The intersection of \( x=a \) and \( y=b \) is always the point \( (a,b) \). Recognize common Pythagorean triples like (3, 4, 5) to quickly calculate distances.

 

Question 10. The points (-4, 0),(4, 0),(0, 3) are the vertices of a
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Answer: (b) isosceles triangle
To determine the type of triangle, we need to calculate the lengths of all three sides using the distance formula.
Let the points be A(-4, 0), B(4, 0), and C(0, 3).
Length of side AB:
\( AB = \sqrt{(4 - (-4))^2 + (0 - 0)^2} \)
\( = \sqrt{(4 + 4)^2 + 0^2} \)
\( = \sqrt{8^2} = \sqrt{64} = 8 \) units.
Length of side BC:
\( BC = \sqrt{(0 - 4)^2 + (3 - 0)^2} \)
\( = \sqrt{(-4)^2 + 3^2} \)
\( = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
Length of side AC:
\( AC = \sqrt{(0 - (-4))^2 + (3 - 0)^2} \)
\( = \sqrt{(0 + 4)^2 + 3^2} \)
\( = \sqrt{4^2 + 3^2} \)
\( = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
Since \( BC = AC = 5 \) units, two sides of the triangle are equal in length.
Therefore, the triangle formed by these points is an isosceles triangle.
In simple words: To find out what kind of triangle these points make, we first measure the length of each of its three sides using the distance formula. If two sides are the same length, it's an isosceles triangle. If all three are the same, it's equilateral. If all are different, it's scalene.

🎯 Exam Tip: Always calculate all three side lengths. Compare them: if all are equal, it's equilateral; if two are equal, it's isosceles; if all are different, it's scalene. You can also check for a right triangle using the Pythagorean theorem \( a^2 + b^2 = c^2 \).