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 I**, **II**, **III**, 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 (x_{1}, y_{1}) and (x_{2}, y_{2}) on the x-y plane, then the distance between two pints P and Q = **((x _{1 }– x_{2})^{2} + (y_{1 }– y_{2})^{2})^{0.5}**

**Section Formula:**

** **If any point C (x, y) divides the line segment joining the points A (x_{1}, y_{1}) and B (x_{2, }y_{2}) in the ratio m: n internally,

X = (mx_{2} + nx_{1})/ (m + n)

Y = (my_{1 }+ my_{2})/ (m + n)

If any point C (x, y) divides the line segment joining the points A (x_{1}, y_{1}) and B (x_{2, }y_{2}) in the ratio m: n externally,

X = (mx_{2} – nx_{1})/ (m + n)

Y = (my_{1 }– my_{2})/ (m + n)

**Slope of a line:**

** **The slope of a line joining two points A (x_{1}, y_{1}) and B (x_{2}, y_{2}) is denoted by m and is given by m = (y_{2 }– y_{1})/ (x_{2 }– x_{1}) = 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 (x_{1}, y_{1}) and having slope m is given by

(y – y_{1}) = m (x – x_{1})

**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 3^{rd} 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.