# 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(x
_{1}, y_{1}) and B(x_{2}, y_{2})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(x
_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) 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(x
_{1}, y_{1}) and C(x_{2}, y_{2}) 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(x
_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) and D(x_{4}, y_{4}) 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(x
_{1}, y_{1}) and B(x_{2}, y_{2}), 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(x
_{1}, y_{1}) and B(x_{2}, y_{2}), then the coordinates of P are given by – $$P = {\LARGE[} \frac{x_1 + x_2}{2} \ , \ \frac{y_1 + y_2}{2} {\LARGE]}$$

- If there are two points A(x
**External Division –**If there are two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}), 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(x
_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) 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(x
_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}), 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(x
_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) 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(x
_{1}, y_{1}) and Q(x_{2}, y_{2}) 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 x_{1}= x_{2}, 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 = \(\frac{Change \ in \ y}{Change \ in \ x}\) .

- Let P(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(x_{1}, y_{1}). 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(x_{1}, y_{1}) is – $$(y \ – \ y_1) = m(x \ – \ x_1)$$**Two-Point Form –**The equation of a non-vertical line “L” passing through P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) 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 (x_{1}, y_{1}) and slope m | Point – Slope Form | $$(y \ – \ y_1) = m(x \ – \ x_1)$$ |

Two Points (x_{1}, y_{1}) and (x_{2}, y_{2}) | If x_{1} = x_{2}, Vertical Line EquationIf x _{1} \( \neq \) x_{2}, 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(a
_{1}x+b_{1}y+c_{1}) and M(a_{2}x+b_{2}y+c_{2}), 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 m
_{1}and m_{2}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 (x
_{1}, y_{1}) 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 a
_{1}x+b_{1}y+c_{1}=0 and a_{2}x+b_{2}y+c_{2}=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 $$x^{2}+y^{2}=r^{2}$$