CAT Quant: Coordinate Geometry – Important Formulas and Concepts

Coordinate Plane

A coordinate system consists of two lines; a horizontal line called the x-axis and vertical line caleed the y-axis both intersecting at a point called the origin (0,0).
Points to the right of the origin on the x-axis have positive x-coordinates and points to the left of the origin on the x-axis have negative x-coordinates.
Points above the origin on the y-axis have positive y-coordinates and points below the origin on the y-axis have negative y-coordinates.
Coordinates of a point are given by an ordered pair (x, y), where x is called the abscissa and y is called the ordinate of the point. X represents the horizontal distance from the y-axis (positive to the right, negative to the left) and Y represents the vertical distance from the x-axis (positive upwards, negative downwards).
The x-axis and y-axis divides the cartesian plane into 4 – Quadrants
- Quadrant 1 – both x-coordinate and y-coordinate is positive ; P(+x, +y).
- Quadrant 2 – x-coordinate is negative and y- coordinate is positive ; P(-x, +y).
- Quadrant 3 – x-coordinate is negative and y-coordinate is negative ; P(-x, -y).
- Quadrant 4 – x-coordinate is positive and y-coordinate is negative ; P(+x, -y).

Distance Formula

The distance between two points A(x₁, y₁) and B(x₂, y₂)is given by – $$AB = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
The distance between the origin O(0, 0) and a point A(x, y) – $$OA = \sqrt{(x + y)}$$

Area of Triangle

The area of the triangle formed by the vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by – $$ar\bigtriangleup ABC = \frac{1}{2} | (x_1y_2 \ -\ x_2y_1) + (x_2y_3 \ – \ x_3y_2) + (x_3y_1 \ – \ x_1y_3)|$$
The area of the triangle formed by the vertices A(0, 0), B(x₁, y₁) and C(x₂, y₂) is given by – $$ar\bigtriangleup ABC = \frac{1}{2} | (x_1y_2 \ – \ x_2y_1)|$$

Area of Quadrilateral

The area of a quadrilateral formed by the points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) and D(x₄, y₄) is given by – $$ar\square ABCD = = \frac{1}{2} | (x_1y_2 \ -\ x_2y_1) + (x_2y_3 \ – \ x_3y_2) + (x_3y_4 \ – \ x_4y_3) + (x_4y_1 \ – \ x_1y_4)|$$

Section Formula

Internal Division –
- If there are two points A(x₁, y₁) and B(x₂, y₂), then the coordinates of the point P, which divides the line joining AB internally in the ratio m:n in given by – $$P = {\LARGE[} \frac{mx_2 + nx_1}{m+n} \ , \ \frac{my_2 + my_1}{m+n} {\LARGE]}$$ Note : The point P is between A and B for internal division.
- If the P is the midpoint of the line joining A(x₁, y₁) and B(x₂, y₂), then the coordinates of P are given by – $$P = {\LARGE[} \frac{x_1 + x_2}{2} \ , \ \frac{y_1 + y_2}{2} {\LARGE]}$$
External Division – If there are two points A(x₁, y₁) and B(x₂, y₂), then the coordinates of the point P, which divides the line joining AB externally in the ratio m:n in given by – $$P = {\LARGE[} \frac{mx_2 \ – \ nx_1}{m-n} \ , \ \frac{my_2 \ – \ my_1}{m-n} {\LARGE]}$$ Note : The point P is beyond A and B for external division.

Centroid

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of a triangle, then the centroid of the triangle (denoted by G) is given by – $$G = {\LARGE[} \frac{x_1 + x_2 + x_3}{3} \ , \ \frac{y_1 + y_2 + y_3}{3} {\LARGE]}$$ Note : The centroid of a triangle is the point of concurrence of the medians of a triangle.

Collinearity

If there are three distinct points in a plane A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃), there may be two possibilities. They may be a triangle or a straight line.
When three or more points form a straight line they are said to be “collinear”.
Any Any of the following conditions are enough to show collinearity of the given three points.
- (AB + BC) = CA or (AC +CB) = AB or (AB + AC) = BC
- The area of the triangle formed by the vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) equals “0”. $$ar\bigtriangleup ABC = \frac{1}{2} | (x_1y_2 \ -\ x_2y_1) + (x_2y_3 \ – \ x_3y_2) + (x_3y_1 \ – \ x_1y_3)| = 0$$

Straight line

We know that there is one and only one line joining two distinct points P and Q from plane geometry
Slope of a Line – The slope of a line is a number that describes both the direction and steepness of a line.
- Let P(x₁, y₁) and Q(x₂, y₂) be two distinct points. The slope “m” of the line “L” containing points P and Q is given by the formula – $$m = \frac{y_2 \ – \ y_1}{x_2 \ – \ x_1} \ , \ if (x_1 \neq x_2)$$ Note : If x₁ = x₂, then slope “m” of line “L” is undefined (since this results in division by 0) and L is a vertical line.
- For a non vertical line , slope = $\frac{Change \ in \ y}{Change \ in \ x}$ .

Equations of Lines

Vertical Lines – The equation of a vertical line passing through a point P(a.0) is given by the equation “x=a” where “a” is a real number.
Non Vertical Line – Let “L” be a non-vertical line with slope m containing A(x₁, y₁). For any point A on line L, we have – $$m = \frac{y \ – \ y_1}{x \ – \ x_1} \ or \ (y \ – \ y_1) = m(x \ – \ x_1)$$
- Point Slope Form – The equation of a non-vertical line “L” with slope “m” and passing through the point A(x₁, y₁) is – $$(y \ – \ y_1) = m(x \ – \ x_1)$$
- Two-Point Form – The equation of a non-vertical line “L” passing through P(x₁, y₁) and Q(x₂, y₂) is – $$(y \ – \ y_1) = \frac{y_2 \ – \ y_1}{x_2 \ – \ x_1} {(x \ – \ x_1)}$$
- General Form – The equation of a line “L” in the general form when it is written as (ax+by+c = 0) where, a, b and c are real numbers with either a $ \neq $ 0 or b $ \neq $ 0.
  - If a = 0 and b $ \neq $ 0, then L will be a horizontal line.
  - If b = 0 and a $ \neq $ 0, then Ll will be a vertical line.
  - If c = 0, then L passes through the origin.

Intercepts

The portion cut off by a line on the coordinate axes is called the intercepts.
The x-intercept is the portion on the x-axis and the y-intercept is the portion on the y-axis. $$Intercept \ Form : \frac{x}{a} + \frac{y}{b} = 1$$ where a-intercept is”a” and y-intercept is “b”.
Slope Intercept form – The equation of a line with slope “m” and y-intercept “b” is given by – $$y = mx +b$$
- When the equation is written in this form, the coefficient of x is the slope and the constant term gives the y-intercept of the line.
- y is explicitly written in terms of x. So this form is also termed as the explicit form of the line.
The following table summaries the various forms of equations of straight line:-

YOU ARE GIVEN	YOU USE	EQUATION
Point (x₁, y₁) and slope m	Point – Slope Form	$$(y \ – \ y_1) = m(x \ – \ x_1)$$
Two Points (x₁, y₁) and (x₂, y₂)	If x₁ = x₂, Vertical Line Equation If x₁ $ \neq $ x₂, Two Point Form	$$x = x_1$$ $$(y \ – \ y_1) = \frac{y_2 \ – \ y_1}{x_2 \ – \ x_1} {(x \ – \ x_1)}$$
x and y intercepts, a and b	Intercept Form	$$\frac{x}{a} + \frac{y}{b} = 1$$
Slope m, y-intercept b	Slope Intercept Form	$$y = mx + b$$

Parallel and Intersecting Lines

Let there be two lines L and M, Exactly one of the three relationships must hold for lines L and M –
- All the points on L are the same as the points on M (Identical Lines).
- L and M have no points in common (Parallel Lines).
- L and M have exactly one point in common (Intersecting Lines).
- For two intersecting lines L(a₁x+b₁y+c₁) and M(a₂x+b₂y+c₂), the point of intersection P is given by – $$P = {\LARGE[} \frac{b_1c_2 \ – \ b_2c_1}{a_1b_2 \ – \ a_2b_1} \ , \ \frac{c_1a_2 \ – \ c_2a_1}{a_1b_2 \ – \ a_2b_1} {\LARGE]}$$

Angle between Two Lines

If m₁ and m₂ are the slopes of two lines, the angle ” $ \theta $” between them is given by – $$tan \theta = {\LARGE |} \frac{m_1 \ – \ m_2}{1 + m_1m_2} {\LARGE |}$$
Condition for Parallel Lines – $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ or \ m_1 = m_2$$
Condition for Perpendicular Lines – $$a_1a_2 = b_1b_2 = 0 \ or \ m_1m_2 = \ -1$$

Additional Points

The general form of the equation of a straight line is ax+by+c=0 Here, the y-intercept is $- \frac{c}{b} $, the x-intercept is $ – \frac{c}{a} $ and the slope is $ – \frac{a}{b} $.
If ax+by+c=0 is the equation of a line, the perpendicular distance from a point (x₁, y₁) to this line is given by – $$ {\LARGE|}\frac{ax_1+by_1+c}{\sqrt{a^{2}+b^{2}}} {\LARGE |}$$
The distance between two parallel straight lines a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0 is given by $$\frac{|c_{1}-c_{2}|}{\sqrt{a^{2}+b^{2}}}$$
The equation of a circle centered at (h,k) with radius ‘r’ units is $$(x-h)^{2}+(y-k)^{2}=r^{2}$$
The equation of a circle centered at the origin with radius ‘r’ units is $$x^{2}+y^{2}=r^{2}$$