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.
- 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}. - Slope of the required line = −1m=−34-\frac{1}{m} = -\frac{3}{4}.
- 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
- Astronomy:
Coordinate geometry is used to calculate distances between planets and map the universe. - Mobile App Development:
Apps like Uber and Google Maps use coordinate systems to plot locations and calculate distances. - Sports and Gaming:
Coordinate geometry helps in analyzing player movements, designing game levels, and calculating trajectories in sports. - Engineering:
Engineers use coordinate geometry in structural design, such as calculating the alignment of beams and supports.
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