GRE: COORDINATE GEOMETRY

Two real number lines that are perpendicular to each other and that intersect at their respective zero points define a rectangular coordinate system. Often called the x-y coordinate system or x-y plane. The horizontal number line is called the x-axis and the vertical number line is called the y-axis. The point where the two axes intersect is called the origin, denoted by O. The positive half of the x-axis is to the right of the origin, and the positive half of the y-axis is above the origin. The two axes divide the plane into four regions called quadrants IIIIII, and IV, as shown in the figure below.

 

Each point P in the x-y plane can be identified with an ordered pair (x, y) of real numbers and is denoted by P (x, y). The first number is called the x-coordinate, and the second number is called the y-coordinate. A point with coordinates (x, y) is located lxl units to the right of the y-axis if x is positive or to the left of the y-axis if x is negative. Also, the point is located lyl units above the x-axis if y is positive or below the x-axis if y is negative. If x = 0, the point lies on the y-axis, and if y = 0, the point lies on the x-axis. The origin has coordinates (0, 0).

In the figure above, the point P (4, 2) is 4 units to the right of the y-axis and 2 units above the x-axis, and the point P’’’ (-4, -2) is 4 units to the left of the y-axis and 2 units below the x-axis.

Distance Formula:

If Two points P and Q are such that they are represented by the points (x1, y1) and (x2, y2) on the x-y plane, then the distance between two pints P and Q = ((x– x2)2 + (y– y2)2)0.5

Section Formula:

 If any point C (x, y) divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n internally,

 

X = (mx2 + nx1)/ (m + n)

Y = (my+ my2)/ (m + n)

If any point C (x, y) divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n externally,

X = (mx2 – nx1)/ (m + n)

Y = (my– my2)/ (m + n)

Slope of a line:

 The slope of a line joining two points A (x1, y1) and B (x2, y2) is denoted by m and is given by m = (y– y1)/ (x– x1) = tan , where  is the angle that the line makes with the positive direction of x-axis.

Equation of line:

Slope Intercept form:

The equation of a straight line passing through the point A (x1, y1) and having slope m is given by

(y – y1) = m (x – x1)

Intercept form:

The equation of a straight line making intercepts a and b on the axes of x and y respectively is given by

x/a + y/b = 1

 

                               

Perpendicularity and parallelism:

Conditions for two lines to be parallel:

Two lines are said to be parallel if their slopes are equal.

Conditions for two lines to be perpendicular:

Two lines are said to be perpendicular if the product of slopes of two lines is equal to -1.

 

Area of triangle:

The area of a triangle whose vertices are \(A({ x }_{ 1 },{ y }_{ 1 }),\quad B({ x }_{ 2 },{ y }_{ 2 }),\quad C({ x }_{ 3 },{ y }_{ 3 })\)  is given by

\(=\frac { 1 }{ 2 } ({ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }))\)

Since the area can’t be negative, we have to take the modulus value given by the above equation.

Problem 1) Find the area of the triangle whose vertices are (1. 3), (-7, 6) & (5, 1)?

solution.  Vertices are A(1,3), B(-7,6) and C(5,-1)

Area of triangle =

\(=\frac { 1 }{ 2 } ({ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }))\) \(=\frac { 1 }{ 2 } (1(6+1)-7(-1-3)+5(3-6))\\ =\frac { 1 }{ 2 } (7-7(-4)+5(-3))\\ =\frac { 1 }{ 2 } (7+28-15)\\ =\frac { 1 }{ 2 } (28-8)\\ =10\quad Sq.\quad units\)

 

Problem 2) What is the area of a triangle with vertices (-2, 4), (2, 4) & (-6, 6) in the coordinate plane?

solution.

 

 

 

 

 

 

 

 

 

 

 

 

 

Let point A(-6, 6), point B(-2, 4) & C(2,4).

Line BC is parallel to x-axis.

Slope of BC = \((\frac { 4-4 }{ 2+2 } )=0\)

We can use line BC as the Base of triangle.

Drop a height vertically from (-6, 6).

Height = different of y coordinates of points A & B = 6-4 = 2

Length of Base (BC) = difference between the \(x\) coordinates of point B & C = 2 – ( -2) = 4

Area of triangle = \(\frac { 1 }{ 2 } \times 4\times 2=4\quad Sq.\quad units\)

 

Problem 3) What will be the reflection of the point (4, 5) in the 3rd quadrant?

solution.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Point (-4,-5) is reflection of the point (4,5) in the third quadrant.

 

Problem 4) If the origin gets shifted to (2, 2), then what will be the new coordinates of the point (4, -2)?

a) (-2, 4)

b) (2, 4)

c) (4, 2)

d) (2, -4)

solution.

 

 

 

 

 

 

 

 

 

 

 

 

 

If origin gets shifted to (2, 2), horizontal distance of point A from y-axis becomes (2) & vertical distance of point. A from x-axis becomes (4) in the downward direction.

Therefore, new co-ordinate of the point will be (2, -4)

 

Problem 5) What will be the equation of the straight line that passes through the intersection of the straight lines \(2x-3y+4=0\) & \(3x+4y-5=0\)  and is perpendicular to the straight lime \(3x-4y-5=0\)?

solution.  Line \(2x-3y+4=0\) & \(3x+4y-5=0\)  intersect at point \((\frac { -1 }{ 17 } ,\frac { 22 }{ 17 } )\)

Slope of straight line \(3x-4y-5=0\) is \(\frac { 3 }{ 4 }\).

Slope of line perpendicular to this line = \(\frac { -4 }{ 3 }\)

Equation of line having slope \(\frac { -4 }{ 3 }\) & passing through the point \((\frac { -1 }{ 17 } ,\frac { 22 }{ 17 } )\) is

\((y-\frac { 22 }{ 17 } )=\frac { -4 }{ 3 } (x+\frac { 1 }{ 17 } )\\ 3y-\frac { 66 }{ 17 } =-4(x+\frac { 1 }{ 17 } )\\ 51y-66=-4(17)(x+\frac { 1 }{ 17 } )\\ 51y-66=-68x-4\\ 68x+51y=-4+66=62\\ Equation={ (68x+5y=62) }^{ Ans }\)

 

Problem 6) Line m & n are perpendicular, neither line is vertical or line m passes through the origin.

Quantity A:         The product of the slopes of lines m & n

Quantity B:         The product of the x-intercepts of lines m & n

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

solution.  The product of slope of perpendicular lines = -1

(The only exception is when one of the lines has an undefined slope because its vertical, but that case has been specifically excluded).

If line m passes through the origin, its x-intercept is 0, So regardless of the x-intercept of line n.

Quantity B is 0.

Therefore, quantity B is greater.

 

Problem 7) Which of the following could be the slope of a line that passes through the point (-2, -3) & crosses the y-axis above the origin?

 

 

 

 

 

 

 

 

 

 

 

 

 

a) \(-\frac { 2 }{ 3 }\)

b) \(\frac { 3 }{ 7 }\)

c) \(\frac { 3 }{ 2 }\)

d) \(\frac { 5 }{ 3 }\)

e) \(\frac { 9 }{ 4 }\)

f) 4

solution.  Slope of the line m passes through the origin & through the point (-2,-3) = \(\frac { -3 }{ -2 } =\frac { 3 }{ 2 }\)

Let angle formed by line m with positive x-axis be \(\theta\), \(\tan { \theta =\frac { 3 }{ 2 } }\)

 

Angle formed by all the lines passing through the point (-2,-3) & crossing the y-axis above the origin lines between \(\theta \quad and\quad { 90 }^{ o }\).

\(\tan { \theta =\frac { 3 }{ 2 } }\) \(\tan { { 90 }^{ 0 }=\infty }\)

Therefore, slope of all lines passing through the point (-2,-3) & crossing the y-axis above the origin lies between \(\frac { 3 }{ 2 } \quad and\quad \infty\).

Therefore, \(\frac { 5 }{ 3 } ,\quad \frac { 9 }{ 4 }\) & 4 could be the slope of such lines.

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