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Trigonometry & Coordinate

Trigonometry And Coordinate

MODULES

Basics of Triangle
Trigonometric Formulae
Quadrants & Graph
Basics of Coordinate Geometry
Equation of a Line
Other Line Properties
Triangle, Quadrilateral & Circle
Past Questions

PRACTICE

Trigonometry & Coordinate : Level 1
Trigonometry & Coordinate : Level 2
Trigonometry & Coordinate : Level 3
ALL MODULES

CAT 2025 Lesson : Trigonometry & Coordinate - Basics of Coordinate Geometry

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3. Coordinate Geometry

In this section we shall look at 2-Dimensional geometry in a coordinate plane. We will use a graph to examine the relationship between geometry and equations. The graph will contain two number lines – the horizontal xxx-axis and the vertical yyy-axis as seen here.



A point in the graph (specifically called the Cartesian Coordinate System) is represented with respect to its distance from the xxx and yyy axis. Every point has a unique pair of number coordinates written as (a,ba, ba,b), where a is the distance from the yyy-axis and b is the distance from the xxx-axis.
Over here, a is also called the x-coordinate or abscissa and b is called the y-coordinate or ordinate. In the graph below are 4 points from the 4 quadrants. Note that the perpendiculars from the points to the axes determine their abscissa and ordinate.



3.1 Distance between points

Distance between two points (x1,y1x_{1}, y_{1}x1​,y1​) and (x2,y2x_{2}, y_{2}x2​,y2​) is (x1−x2)2+(y1−y2)2\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}(x1​−x2​)2+(y1​−y2​)2​.

(Note that distance is always a positive value, therefore the negative result of the square root can be ignored.)

Example 8

What is the distance between the points (−2,5-2, 5−2,5) and (−7,−7-7, -7−7,−7)?

Solution

Distance = (−2−(−7))2+(5−(−7))2=52+122=169=13\sqrt{(-2 - (-7))^{2} + (5 - (-7))^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{169} = 13(−2−(−7))2+(5−(−7))2​=52+122​=169​=13

Answer: 131313

3.2 Internal and External Division of a Segment

Internal Division: Where the line segment joining two points A and B with coordinates (x1,y1x_{1}, y_{1}x1​,y1​) and (x2,y2x_{2}, y_{2}x2​,y2​) needs to be divided internally by a point P (x3,y3x_{3}, y_{3}x3​,y3​) such that AP : PB is in the ratio m:nm : nm:n, then the coordinates of P,
i.e., (x3,y3)=(mx2+nx1m+n,my2+ny1m+n)(x_{3}, y_{3}) = \left(\dfrac{mx_{2} + nx_{1}}{m + n}, \dfrac{my_{2} + ny_{1}}{m + n} \right)(x3​,y3​)=(m+nmx2​+nx1​​,m+nmy2​+ny1​​)



External Division: Where the line segment joining two points A and B with coordinates (x1,y1x_{1}, y_{1}x1​,y1​) and (x2,y2x_{2}, y_{2}x2​,y2​) need to be divided externally by a point P (x3,y3x_{3}, y_{3}x3​,y3​) such that AP : PB is in the ratio m:nm : nm:n (where m>nm > nm>n), then the coordinates of P,
i.e., (x3,y3)=(mx2−nx1m−n,my2−ny1m−n)(x_{3}, y_{3}) = \left(\dfrac{mx_{2} - nx_{1}}{m - n}, \dfrac{my_{2} - ny_{1}}{m - n} \right)(x3​,y3​)=(m−nmx2​−nx1​​,m−nmy2​−ny1​​)



Note that the point should lie beyond B, so that m>nm > nm>n. If m<nm < nm<n, then B will lie before A as BP > AP. In either case, the formula remains the same.

Midpoint of a Segment: The midpoint divides a line segment in the ratio of 1 : 1. Therefore, we need to substitute m=1m = 1m=1 and n=1n =1n=1 in the above formula.

Midpoint (x3,y3)=(x1+x22,y1+y22)(x_{3},y_{3}) = \left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2} \right)(x3​,y3​)=(2x1​+x2​​,2y1​+y2​​)

Example 9

Points A, B and P are collinear. The coordinates of A are (2, 5) and B are (-3, 7). Find the coordinates of point P if P is the
(i) is the Midpoint      (ii) lies on AB and AP : PB = 2 : 5
(iii) does not lie on AB and AP : PB = 2 : 5

Solution

(i) P is the Midpoint

Let A (2,52, 52,5) be (x1,y1x_{1}, y_{1}x1​,y1​), B (−3,7-3, 7−3,7) be (x2,y2x_{2}, y_{2}x2​,y2​) and P be (x3,y3x_{3}, y_{3}x3​,y3​)

Midpoint P (x3,y3)=(x1+x22,y1+y22)(x_{3}, y_{3}) = \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)(x3​,y3​)=(2x1​+x2​​,2y1​+y2​​)

⇒ P(x3,y3)=(2−32,5+72)\mathrm{P}(x_3, y_3) = \left(\dfrac{2 - 3}{2}, \dfrac{5 + 7}{2} \right)P(x3​,y3​)=(22−3​,25+7​)

⇒ P(x3,y3)=(−12,6)\mathrm{P}(x_3, y_3) = \left(\dfrac{-1}{2}, 6 \right)P(x3​,y3​)=(2−1​,6)

(ii) P lies on AB and AP : PB = 2:52 : 52:5
Since P (x3,y3x_3, y_3x3​,y3​) divides A (x1,y1x_1, y_1x1​,y1​) and B (x2,y2x_2, y_2x2​,y2​) in the ratio m:nm : nm:n, p(x3,y3)=(mx2+nx1m+n,my2+ny1m+n)\mathrm{p}(x_3, y_3) = \left(\dfrac{mx_{2} + nx_{1}}{m + n}, \dfrac{my_{2} + ny_{1}}{m + n} \right)p(x3​,y3​)=(m+nmx2​+nx1​​,m+nmy2​+ny1​​)

Let AP : PB be m:n=2:5m : n = 2 : 5m:n=2:5

⇒ P(x3,y3)=(2(−3)+5(2)2+5,2(7)+5(5)2+5)\mathrm{P}(x_3, y_3) = \left(\dfrac{2(-3) + 5(2)}{2 + 5}, \dfrac{2(7) + 5(5)}{2 + 5} \right)P(x3​,y3​)=(2+52(−3)+5(2)​,2+52(7)+5(5)​)

⇒ P(x3,y3)=(−6+107,14+257)\mathrm{P}(x_3, y_3) = \left(\dfrac{-6 + 10}{7}, \dfrac{14 + 25}{7} \right)P(x3​,y3​)=(7−6+10​,714+25​)

⇒ P(x3,y3)=(47,397)\mathrm{P}(x_{3}, y_3) = \left(\dfrac{4}{7}, \dfrac{39}{7} \right)P(x3​,y3​)=(74​,739​)

(iii) P does not lie on AB and AP : PB = 2:52 : 52:5

Since P (x3,y3x_3, y_3x3​,y3​) divides A (x1,y1x_1, y_1x1​,y1​) and B (x2,y2x_2, y_2x2​,y2​) externally in the ratio m:nm : nm:n, where m>nm > nm>n m:n=5:2m : n = 5 : 2m:n=5:2

⇒ P(x3,y3)=(mx2−nx1m−n,my2−ny1m−n)\mathrm{P}(x_3, y_3) = \left(\dfrac{mx_{2} - nx_{1}}{m - n}, \dfrac{my_{2} - ny_{1}}{m - n}\right)P(x3​,y3​)=(m−nmx2​−nx1​​,m−nmy2​−ny1​​)

⇒ P(x3,y3)=(5(−3)−2(2)5−2,5(7)−2(5)5−2)\mathrm{P}(x_3, y_3) = \left(\dfrac{5(-3) - 2(2)}{5 - 2}, \dfrac{5(7) - 2(5)}{5 - 2}\right)P(x3​,y3​)=(5−25(−3)−2(2)​,5−25(7)−2(5)​)

⇒ P(x3,y3)=(−15−43,35−103)\mathrm{P}(x_{3}, y_{3}) = \left(\dfrac{-15 - 4}{3}, \dfrac{35 - 10}{3} \right)P(x3​,y3​)=(3−15−4​,335−10​)

⇒ P(x3,y3)=(−193,253)\mathrm{P}(x_{3}, y_{3}) = \left(\dfrac{-19}{3}, \dfrac{25}{3} \right)P(x3​,y3​)=(3−19​,325​)

Answer: (i) (−12,6)\left(\dfrac{-1}{2}, 6 \right)(2−1​,6), (ii) (47,397)\left(\dfrac{4}{7},\dfrac{39}{7} \right)(74​,739​) , (iii) (−193,253)\left(\dfrac{-19}{3}, \dfrac{25}{3} \right)(3−19​,325​)

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