Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 7 Set 7.2 Co ordinate Solutions

Question 1. On a graph paper plot the points A(3, 0), B(3, 3), C(0, 3). Join A, B and B, C. What is the figure formed? Soiution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर बिंदु A(3,0), B(3,3) और C(0,3) को दर्शाता है। बिंदु O(0,0) (मूल बिंदु) को भी दिखाया गया है। रेखाखंड OA, AB, BC, और CO को जोड़कर एक आकृति बनाई गई है, जो OABC एक वर्ग बनाती है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है।
Answer: d(O, A) = 3 cm, d(A, B) = 3 cm, d(B, C) = 3 cm, d(O, C) = 3 cm and each angle of OABC is 90°
∴ OABC is a square.
In simple words: When points A(3,0), B(3,3), and C(0,3) are plotted on a graph and joined with the origin O(0,0), the resulting figure OABC is a square because all its sides are equal and all its angles are 90 degrees.

🎯 Exam Tip: When identifying geometric figures formed by points, always check the lengths of sides and the angles between them. For a square, all four sides must be equal and all angles must be 90 degrees.

Question 2. Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its left.
Answer: Solution: The equation of a line parallel to the Y-axis is x = a.
Since, the line is at a distance of 7 units to the left of Y-axis,
∴ a = -7
∴ x = -1 is the equation of the required line.
In simple words: A line parallel to the Y-axis always has the equation x=a. Since it's 7 units to the left, 'a' becomes -7, so the equation is x = -7 (although the provided solution states x = -1, which is a calculation error in the original text).

🎯 Exam Tip: Remember that lines parallel to the Y-axis have equations of the form x=a, and lines parallel to the X-axis have equations of the form y=b. "Left" implies negative 'a', and "below" implies negative 'b'.

Question 3. Write the equation of the line parallel to the X-axis at a distance of 5 units from it and below the X-axis.
Answer: Solution: The equation of a line parallel to the X-axis is y = b.
Since, the line is at a distance of 5 units below the X-axis.
∴ b = -5
∴ y = -5 is the equation of the required line.
In simple words: A line parallel to the X-axis always has the equation y=b. Since it's 5 units below the X-axis, 'b' becomes -5, so the equation is y = -5.

🎯 Exam Tip: Positive distances are above the X-axis or to the right of the Y-axis; negative distances are below the X-axis or to the left of the Y-axis.

Question 4. The point Q( -3, -2) lies on a line parallel to the Y-axis. Write the equation of the line and draw its graph.
Answer: Solution: The equation of a line parallel to the Y-axis is x = a.
Here, a = -3
∴ x = -3 is the equation of the required line.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर Y-अक्ष के समानांतर रेखा x = -3 को दर्शाता है। बिंदु Q(-3, -2) इस रेखा पर स्थित है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है।
In simple words: If a point Q(-3, -2) lies on a line parallel to the Y-axis, then its x-coordinate (-3) determines the equation of the line, which is x = -3.

🎯 Exam Tip: For a line parallel to the Y-axis, all points on it share the same x-coordinate. For a line parallel to the X-axis, all points share the same y-coordinate.

Question 5. Y-axis and line x = 4 are parallel lines. What is the distance between them?
Answer: Solution: Equation of Y-axis is x = 0.
Equation of the line parallel to the Y-axis is x = - 4. ... [Given]
∴ Distance between the Y-axis and the line x = - 4 is 0 – (-4) ... [0 > -4]
= 0 + 4 = 4 units
∴ The distance between the Y-axis and the line x = 4 is 4 units.
[Note: The question is modified as X-axis cannot be parallel to the line x = - 4.]
In simple words: The Y-axis is represented by x=0. The distance between x=0 and x=-4 (or x=4 as stated in the solution's conclusion) is found by taking the absolute difference of their x-coordinates, which is 4 units.

🎯 Exam Tip: The distance between two vertical lines \(x = a_1\) and \(x = a_2\) is \(|a_1 - a_2|\). Similarly, for horizontal lines \(y = b_1\) and \(y = b_2\), the distance is \(|b_1 - b_2|\).

Question 6. Which of the equations given below have graphs parallel to the X-axis, and which ones have graphs parallel to the Y-axis? [1 Mark each]
(i) x = 3
(ii) y - 2 = 0
(iii) x + 6 = 0
(iv) y = -5
Answer: Solution:
(i) The equation of a line parallel to the Y-axis is x = a.
∴ The line x = 3 is parallel to the Y-axis.
(ii) y - 2 = 0
∴ y = 2
The equation of a line parallel to the X-axis is y = b.
∴ The line y - 2 = 0 is parallel to the X-axis.
(iii) x + 6 = 0
∴ x = -6
The equation of a line parallel to the Y-axis is x = a.
∴ The line x + 6 = 0 is parallel to the Y-axis.
(iv) The equation of a line parallel to the X-axis is y = b.
∴ The line y = - 5 is parallel to the X-axis.
In simple words: Equations of the form x=constant are parallel to the Y-axis, while equations of the form y=constant are parallel to the X-axis.

🎯 Exam Tip: Identify the variable present in the equation. If only 'x' is present, it's parallel to the Y-axis. If only 'y' is present, it's parallel to the X-axis. Always simplify the equation to the standard form (e.g., y=constant or x=constant) first.

Question 7. On a graph paper, plot the points A(2, 3), B(6, -1) and C(0, 5). If these points are collinear, then draw the line which includes them. Write the co-ordinates of the points at which the line intersects the X-axis and the Y-axis.
Answer: Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस ग्राफ पर बिंदु A(2,3), C(0,5), D(5,0) और B(6,-1) प्लॉट किए गए हैं। ये सभी बिंदु एक सीधी रेखा पर स्थित हैं, जो दर्शाती है कि वे संरेख हैं। यह रेखा X-अक्ष को बिंदु D(5,0) पर और Y-अक्ष को बिंदु C(0,5) पर काटती है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है।
From the graph, the line drawn intersects the X-axis at D(5, 0) and the Y-axis at C(0, 5).
In simple words: Plotting the given points A(2,3), B(6,-1), and C(0,5) on a graph reveals they lie on a single straight line. This line intersects the X-axis at (5,0) and the Y-axis at (0,5).

🎯 Exam Tip: To check for collinearity, plot the points accurately on a graph. If they form a straight line, they are collinear. Always clearly label the intersection points with the axes.

Question 8. Draw the graphs of the following equations on the same system of co-ordinates. Write the co-ordinates of their points of intersection.
x + 4 = 0,
y - 1 = 0,
2x + 3 = 0,
3y - 15 = 0
Answer: Solution:
(i) x + 4 = 0
∴ x = - 4
(ii) y - 1 = 0
∴ y = 1
(iii) 2x + 3 = 0
∴ 2x = -3
∴ x = \( \frac{-3}{2} \)
∴ x = -1.5
(iv) 3y- 15 = 0
3y = 15
y = \( \frac{15}{3} \)
∴ y = 5
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर चार रेखाओं - x+4=0 (एक ऊर्ध्वाधर रेखा), y-1=0 (एक क्षैतिज रेखा), 2x+3=0 (एक ऊर्ध्वाधर रेखा), और 3y-15=0 (एक क्षैतिज रेखा) - के ग्राफ को दर्शाता है। इन रेखाओं के प्रतिच्छेदन बिंदुओं को A(-4,1), B(-1.5,1), C(-1.5,5) और D(-4,5) के रूप में लेबल किया गया है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है।
The co-ordinates of the point of intersection of x + 4 = 0 and y - 1 = 0 are A(-4, 1).
The co-ordinates of the point of intersection of y - 1 = 0 and 2x + 3 = 0 are B(-1.5, 1).
The co-ordinates of the point of intersection of 3y - 15 = 0 and 2x + 3 = 0 are C(-1.5, 5).
The co-ordinates of the point of intersection of x + 4 = 0 and 3y - 15 = 0 are D(-4, 5).
In simple words: After simplifying the equations to x=-4, y=1, x=-1.5, and y=5, plot these four lines. Then, identify and list the coordinates of the four points where these lines intersect each other.

🎯 Exam Tip: When plotting multiple linear equations, ensure each line is correctly drawn based on its simplified form. Points of intersection are critical and must be read accurately from the graph.

Question 9. Draw the graphs of the equations given below.
(i) x + y = 2
(ii) 3x - y = 0
(iii) 2x + y = 1
Answer: Solution:
(i) x + y = 2
∴ y = 2 - x
When x = 0,
y = 2-x
= 2-0
= 2
When x = 1,
y = 2-x
= 2-1
= 1
When x = 2,
y = 2-x
= 0

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर समीकरण x+y=2 का ग्राफ दर्शाता है। रेखा बिंदु (0,2), (1,1) और (2,0) से होकर गुजरती है, जैसा कि गणना की गई तालिका में दिखाया गया है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है।
(ii) 3x - y = 0
∴ y = 3x
When x = 0,
y = 3x
= 3(0)
= 0
When x = 1,
y = 3x
= 3(1)
= 3
When x = -1,
y = 3x
= 3(-1)
= -3

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर समीकरण 3x-y=0 का ग्राफ दर्शाता है। रेखा बिंदु (0,0), (1,3) और (-1,-3) से होकर गुजरती है, जैसा कि गणना की गई तालिका में दिखाया गया है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है।
(iii) 2x + y = 1
∴ y = 1 - 2x
When x = 0,
y = 1 - 2x
= 1 - 2(0)
= 1-0
When x = 1,
y = 1 - 2x
= 1-2(1)
= 1-2
= -1
When x = -1,
y = 1 - 2x
= 1 - 2(-1)
= 1 + 2
= 3

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर समीकरण 2x+y=1 का ग्राफ दर्शाता है। रेखा बिंदु (0,1), (1,-1) और (-1,3) से होकर गुजरती है, जैसा कि गणना की गई तालिका में दिखाया गया है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है।
In simple words: To draw the graph of a linear equation, first find at least three coordinate points (x,y) that satisfy the equation. Plot these points on the graph paper and then draw a straight line passing through them.

🎯 Exam Tip: For drawing linear graphs, calculating at least three points is recommended to ensure accuracy. Always label your axes, scale, and the equation of the line on the graph.

Maharashtra Board Class 9 Maths Chapter 7 Co-Ordinate Geometry Practice Set 7.2 Intext Questions And Activities

Question 1.
(i) Can we draw a line parallel to the X-axis at a distance of 6 units from it and below the X-axis?
(ii) Will all of the points (-3,-6), (10,-6), (\( \frac{1}{2} \), -6) be on that line?
(iii) What would be the equation of this line?(Textbook pg. no. 94)
Answer: Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर एक क्षैतिज रेखा y = -6 को दर्शाता है। यह रेखा X-अक्ष के समानांतर है और X-अक्ष से 6 इकाई नीचे है। इस रेखा पर (-3,-6), (0,-6), (10,-6), और (\( \frac{1}{2} \), -6) जैसे विभिन्न बिंदु दिखाए गए हैं।
(i) Yes.
This line will pass through the point (0,-6).
(ii) Yes.
Here, y co-ordinate of the points (-3, -6), (10,-6), (\( \frac{1}{2} \), -6) is the same, which is -6.
∴ All the above points lie on the same line.
(iii) Since, the line is at a distance of 6 units below the X-axis.
∴ b = -6
∴ Equation of the line is y = -6.
In simple words: Yes, a line parallel to the X-axis and 6 units below it can be drawn; its equation would be y=-6. All points with a y-coordinate of -6 will lie on this line.

🎯 Exam Tip: Understand that all points on a horizontal line have the same y-coordinate, and all points on a vertical line have the same x-coordinate. This is key to writing their equations.

Question 2.
(i) Can we draw a line parallel to the Y – axis at a distance of 2 units from it and to its right?
(ii) Will all of the points (2, 10), (2, 8), (2, -) be on that line?
(iii) What would be the equation of this line? (Textbook pg. no. 95)
Answer: Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर एक ऊर्ध्वाधर रेखा x = 2 को दर्शाता है। यह रेखा Y-अक्ष के समानांतर है और Y-अक्ष से 2 इकाई दाईं ओर है। इस रेखा पर (2,10), (2,8), (2,0), और (2, \( -\frac{1}{2} \)) जैसे विभिन्न बिंदु दिखाए गए हैं।
(i) Yes.
(2, 10)
This line will pass through the point (2, 0).
(2,8)
(ii) Yes.
Here, x co-ordinate of the points (2, 10), (2, 8), (2, \( -\frac{1}{2} \)) is the same, which is 2.
∴ All the above points lie on the same line.
(iii) Since, the line is at a distance of 2 units to the right of Y-axis.
a = 2
∴ Equation of the line is x = 2.
In simple words: Yes, a line parallel to the Y-axis and 2 units to its right can be drawn; its equation would be x=2. All points with an x-coordinate of 2 will lie on this line.

🎯 Exam Tip: The x-coordinate determines the position of a vertical line. If a line is 'a' units to the right of the Y-axis, its equation is x=a. If it's to the left, x=-a.

Question 3. On a graph paper, plot the points (0, 1), (1, 3), (2, 5). Are they collinear? If so, draw the line that passes through them.
(i) Through which quadrants does this line pass ?
(ii) Write the co-ordinates of the point at which it intersects the Y-axis.
(iii) Show any point in the third quadrant which lies on this line. Write the co-ordinates of the point. (Textbook pg. no. 96)
Answer: Solution:
(i) The line passes through the quadrants I, II and III.
(ii) The line intersects the Y-axis at (0, 1).
(iii) (-1,-1)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ पेपर पर बिंदु (0,1), (1,3), (2,5) और (-1,-1) को दर्शाता है। इन बिंदुओं से होकर एक सीधी रेखा खींची गई है, जो यह पुष्टि करती है कि बिंदु संरेख हैं। यह रेखा प्रथम, द्वितीय और तृतीय चतुर्थांश से होकर गुजरती है, और Y-अक्ष को (0,1) पर काटती है। ग्राफ पर 1 सेमी = 1 इकाई का पैमाना उपयोग किया गया है।
In simple words: When points (0,1), (1,3), and (2,5) are plotted, they form a straight line, making them collinear. This line passes through Quadrants I, II, and III, intersects the Y-axis at (0,1), and includes a point like (-1,-1) in the third quadrant.

🎯 Exam Tip: Plotting points accurately is crucial for determining collinearity and identifying intersections with axes. Remember the signs of coordinates in each quadrant to correctly identify points in specific quadrants.