Tag: 9TH MATHS CBSE

Advanced Problem-Solving in Coordinate Geometry

1. Intersection of Lines

Problem: Find the point of intersection of the lines represented by:

• 2x+3y=132x + 3y = 13
• x−2y=−5x – 2y = -5.

Solution:
To find the intersection, solve the two equations simultaneously.

• Equation 1: 2x+3y=132x + 3y = 13
• Equation 2: x−2y=−5x – 2y = -5.

From Equation 2: x=2y−5.x = 2y – 5.

Substitute x=2y−5x = 2y – 5 into Equation 1: 2(2y−5)+3y=13.2(2y – 5) + 3y = 13. 4y−10+3y=13.4y – 10 + 3y = 13. 7y=23.7y = 23. y=237.y = \frac{23}{7}.

Substitute y=237y = \frac{23}{7} into x=2y−5x = 2y – 5: x=2(237)−5.x = 2\left(\frac{23}{7}\right) – 5. x=467−357=117.x = \frac{46}{7} – \frac{35}{7} = \frac{11}{7}.

Answer: The lines intersect at (117,237)\left(\frac{11}{7}, \frac{23}{7}\right).

2. Verifying Collinearity

Problem: Check if the points A(1,2)A(1, 2), B(3,6)B(3, 6), and C(5,10)C(5, 10) are collinear.

Solution:
To verify collinearity, calculate the area of the triangle formed by these points using the formula: Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|.

Substitute A(1,2)A(1, 2), B(3,6)B(3, 6), C(5,10)C(5, 10): Area=12∣1(6−10)+3(10−2)+5(2−6)∣.\text{Area} = \frac{1}{2} \left| 1(6 – 10) + 3(10 – 2) + 5(2 – 6) \right|. =12∣1(−4)+3(8)+5(−4)∣.= \frac{1}{2} \left| 1(-4) + 3(8) + 5(-4) \right|. =12∣−4+24−20∣.= \frac{1}{2} \left| -4 + 24 – 20 \right|. =12∣0∣=0.= \frac{1}{2} \left| 0 \right| = 0.

Since the area is 00, the points are collinear.

Answer: AA, BB, and CC are collinear.

3. Shortest Distance from a Point to a Line

Problem: Find the shortest distance from the point P(4,1)P(4, 1) to the line 3x−4y+5=03x – 4y + 5 = 0.

Solution:
The formula for the shortest distance from a point (x1,y1)(x_1, y_1) to a line Ax+By+C=0Ax + By + C = 0 is: Distance=∣Ax1+By1+C∣A2+B2.\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}.

Here, A=3A = 3, B=−4B = -4, C=5C = 5, x1=4x_1 = 4, y1=1y_1 = 1: Distance=∣3(4)−4(1)+5∣32+(−4)2.\text{Distance} = \frac{|3(4) – 4(1) + 5|}{\sqrt{3^2 + (-4)^2}}. =∣12−4+5∣9+16.= \frac{|12 – 4 + 5|}{\sqrt{9 + 16}}. =∣13∣25=135.= \frac{|13|}{\sqrt{25}} = \frac{13}{5}.

Answer: The shortest distance is 135\frac{13}{5} or 2.6 units2.6 \, \text{units}.

4. Equation of a Line

Problem: Find the equation of a line passing through (2,3)(2, 3) and perpendicular to the line 4x−3y+7=04x – 3y + 7 = 0.

Solution:
For a line perpendicular to another line, the slope of the new line is the negative reciprocal of the given line’s slope.

1. Find the slope of 4x−3y+7=04x – 3y + 7 = 0:
   Rewrite in slope-intercept form (y=mx+cy = mx + c): 3y=4x+7⇒y=43x+73.3y = 4x + 7 \quad \Rightarrow \quad y = \frac{4}{3}x + \frac{7}{3}. Slope (mm) of the given line = 43\frac{4}{3}.
2. Slope of the required line = −1m=−34-\frac{1}{m} = -\frac{3}{4}.
3. Use the point-slope form to find the equation of the line: y−y1=m(x−x1).y – y_1 = m(x – x_1). Substitute (x1,y1)=(2,3)(x_1, y_1) = (2, 3), m=−34m = -\frac{3}{4}: y−3=−34(x−2).y – 3 = -\frac{3}{4}(x – 2). y−3=−34x+64.y – 3 = -\frac{3}{4}x + \frac{6}{4}. y=−34x+32+3.y = -\frac{3}{4}x + \frac{3}{2} + 3. y=−34x+92.y = -\frac{3}{4}x + \frac{9}{2}.

Answer: Equation of the line: y=−34x+92y = -\frac{3}{4}x + \frac{9}{2}.

Applications in Depth

1. Astronomy:
   Coordinate geometry is used to calculate distances between planets and map the universe.
2. Mobile App Development:
   Apps like Uber and Google Maps use coordinate systems to plot locations and calculate distances.
3. Sports and Gaming:
   Coordinate geometry helps in analyzing player movements, designing game levels, and calculating trajectories in sports.
4. Engineering:
   Engineers use coordinate geometry in structural design, such as calculating the alignment of beams and supports.

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Key Concepts

1. Cartesian System

The Cartesian system uses two perpendicular axes:

• The horizontal axis (xx-axis).
• The vertical axis (yy-axis).

Origin (OO): Intersection point of the axes (0,0)(0, 0).
Points are represented as ordered pairs (x,y)(x, y), where:

• xx is the abscissa (horizontal distance from the origin).
• yy is the ordinate (vertical distance from the origin).

2. Quadrants

The plane is divided into four quadrants:

• Quadrant I: x>0,y>0x > 0, y > 0.
• Quadrant II: x<0,y>0x < 0, y > 0.
• Quadrant III: x<0,y<0x < 0, y < 0.
• Quadrant IV: x>0,y<0x > 0, y < 0.

3. Plotting Points

• Start at the origin.
• Move horizontally to the xx-coordinate.
• Move vertically to the yy-coordinate.

Example: Plot (2,3)(2, 3).

• Move 2 units right (positive xx).
• Move 3 units up (positive yy).

4. Distance Formula

The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is: d=(x2−x1)2+(y2−y1)2.d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}.

Example: Find the distance between A(1,2)A(1, 2) and B(4,6)B(4, 6). d=(4−1)2+(6−2)2=32+42=9+16=25=5.d = \sqrt{(4 – 1)^2 + (6 – 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

5. Section Formula

A point P(x,y)P(x, y) dividing a line segment joining (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) in the ratio m:nm:n is: P(x,y)=(mx2+nx1m+n,my2+ny1m+n).P\left(x, y\right) = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right).

Example: Find the point dividing the line joining A(1,2)A(1, 2) and B(4,6)B(4, 6) in the ratio 2:12:1. x=2(4)+1(1)2+1=8+13=3.x = \frac{2(4) + 1(1)}{2 + 1} = \frac{8 + 1}{3} = 3. y=2(6)+1(2)2+1=12+23=4.67.y = \frac{2(6) + 1(2)}{2 + 1} = \frac{12 + 2}{3} = 4.67.

Point P=(3,4.67)P = (3, 4.67).

6. Midpoint Formula

The midpoint of a line segment joining (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is: M(x,y)=(x1+x22,y1+y22).M\left(x, y\right) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).

Example: Find the midpoint of A(1,2)A(1, 2) and B(4,6)B(4, 6). x=1+42=2.5, y=2+62=4.x = \frac{1 + 4}{2} = 2.5, \, y = \frac{2 + 6}{2} = 4.

Midpoint M=(2.5,4)M = (2.5, 4).

7. Area of a Triangle

The area of a triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3) is: Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|.

Example: Find the area of a triangle with vertices A(1,2)A(1, 2), B(4,6)B(4, 6), and C(3,5)C(3, 5). Area=12∣1(6−5)+4(5−2)+3(2−6)∣.\text{Area} = \frac{1}{2} \left| 1(6 – 5) + 4(5 – 2) + 3(2 – 6) \right|. =12∣1(1)+4(3)+3(−4)∣.= \frac{1}{2} \left| 1(1) + 4(3) + 3(-4) \right|. =12∣1+12−12∣=12∣1∣=0.5.= \frac{1}{2} \left| 1 + 12 – 12 \right| = \frac{1}{2} \left| 1 \right| = 0.5.

Answer: Area = 0.5 square units0.5 \, \text{square units}.

Critical Analysis

1. Conceptual Clarity:
   • Understanding quadrants is essential for correctly identifying the position of points.
   • Mastery of formulas like distance, section, and midpoint helps solve geometry problems efficiently.
2. Applications:
   • Coordinate geometry has real-world applications in navigation, engineering, and computer graphics.
   • It lays the foundation for advanced topics like vectors and 3D geometry.
3. Common Mistakes:
   • Confusing the signs of coordinates in different quadrants.
   • Misapplying formulas, especially in the section formula with incorrect ratios.
4. Visualization:
   • Drawing rough sketches improves accuracy.
   • Graphical representation helps in better understanding spatial relationships.

Applications of Coordinate Geometry

1. Map Reading

• Identifying locations on a map using coordinates.

2. Navigation

• GPS systems use coordinate geometry to calculate routes and distances.

3. Design and Animation

• Used in computer-aided design (CAD) and animation software to position objects.

4. Physics

• Analyzing motion in two dimensions.

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