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# Geometry

Coordinate Geometry - 1

__Coordinate Geometry:__

Coordinate Geometry is the branch of mathematics that is a combination of geometry and algebra. In geometry we study about geometrical figures while in algebra we study about equations. So, in coordinate geometry we study about geometrical objects with the help of equations. It was conceived by the great French philosopher and mathematicians Rene Descartes and is widely used trigonometry, calculus, statistics and physics. The coordinates of a place are determined by its latitudes and longitudes. But why do we need to study it?

Consider a plane and a point P on it. Let’s say it is at a distance of 10 units from the line X, but there might be infinite number of points at a distance of 10 units from the line X as shown in the figure.

But if we say it is also at a distance of 15 units from the line Y, then there must be only a unique point at a distance of 10 from X and at a distance of 15 from Y simultaneously.

So, the idea here is that we need to perpendicular lines and the distance from these two lines will give unique coordinates of any point.

__Coordinate Plane:__

Coordinate Plane is an infinite plane in which two perpendicular lines are drawn known as X axis and Y axis respectively and their point of intersection is known as the origin. The two lines divide the coordinate plane into four parts known as the quadrants.

X coordinate or abscissa is the distance of the point from the Y axis and Y coordinate or ordinate is the distance of the point from the X axis. By convention,

Quadrants |
X coordinate |
Y coordinate |

First |
Positive |
Positive |

Second |
Negative |
Positive |

Third |
Negative |
Negative |

Fourth |
Positive |
Negative |

__Distance between two points:__

If P(x_{1},y_{1}) and Q(x_{2},y_{2}) are two points on the Cartesian Plane, then the distance between P and Q is given by:

__Sectional Formula:__

__Internal Division: __

If P(x,y) divides AB in the ratio of m:n internally where the coordinates of A are (x_{1},y_{1}) and the coordinates of B are(x_{2},y_{2}) then,

__Coordinates of Centroid of a Triangle:__

If A(x_{1},y_{1}), B(x_{2},y_{2}) and C(x_{3},y_{3}) are the coordinates of the three vertices of a triangle, then the coordinates of the centroid G, are given by

where centroid is the point of intersection of the medians of a triangle.

__Area of a Triangle:__

The area of a triangle with vertices (x_{1}, y_{1}), (x_{2},y_{2}) and (x_{3},y_{3}) is given by :

**Δ = ½ |[x _{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]|**

**Example 1:** A(2,1), B(3,2) and C(3,0) are the vertices of a triangle. Then, ΔABC is-

1. Equilateral Triangle

2. Isosceles Triangle

3. Isosceles Right Triangle

4. Scalene Triangle

**Solution: **

The triangle is clearly isosceles as AC = BC. Also we see that AC^{2} + BC^{2} = AB^{2} . Thus, the triangle is an isosceles right triangle.

**Option (3) is correct. **

**Example 2:** The opposite vertices of a square are (1,-5) and (6,7). The area of the square is-

**Solution:**

**Example 3:** In what ratio does the point (4/5, 18/5) divide the line joining the points (2,4) and (-1,3)?

**Answer**: Let the point (4/5, 18/5) divide the line segment in the ratio of k : 1

So, 4/5 = (-1*k+2*1)/(k+1)

4(k+1) = 5(2-k)

4k +4 = 10 -5k

9k = 6 or k = 2/3

Thus, the point (4/5, 18/5) divides the line joining the points (2,4) and (-1,3) in the ratio of **2:3 internally.**

**Example 4:** A(2,-1) , B(4,2) and C(3,2) are the three vertices of a triangle ABC. D is the midpoint of AB, E is the midpoint of AC and F is the midpoint of BC. What is the area of ΔDEF?

**Answer: **

As D, E and F are the midpoints of the sides, so

Area of ΔDEF = ¼ (Area of ΔABC)

Area of a triangle is calculated using the formula

Δ = ½ |[x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]|

Hence Area of ΔABC = ½ |[2(2-2) + 4{2-(-1)} + 3(-1 -2)]

= 3/2

So, **Area of ΔDEF = 3/8 units**.

**Example 5:** If the point of intersection of the lines 3x+2y+6 = 0 and 4x-y+8 =0 lies on the line 5x+y+k = 0, then the value of k is?

**Answer:** Solving the equations 3x + 2y + 10 = 0 and 4x – y + 20 = 0 we get the point of intersection as (-2,0).

This point lies on the line 5x +y + k = 0,

So, 5(-2) +0 +k = 0

**Or k = 10**

__Slope of a Line:__

** Angle between two lines: **

If m_{1} and m_{2} are the slopes of two intersecting lines L_{1} and L_{2},then the angle between them is given by:

**Parallel lines:**

If the two lines are parallel, then the angle between them is 0^{0}, so tan0 = 0 and hence m_{1 }=m_{2}, i.e. their slopes are equal.

**Perpendicular Lines.**

If two lines are perpendicular then the angle between them is 90^{0}, and so 1+m_{1}m_{2} = 0 or m_{1}m_{2} = -1

__Straight Line: __

The general form a straight line is ax + by + c = 0. Now there are various ways or formsin which the equation of the line can be determined.

- Slope Intercept Form:

The equation of a line with slope m and intercept on Y – axis equal to c, is **y = mx + c**

**2. Point-Slope Form:**

The equation of the line passing through (x_{1}, y_{1}) and having slope m is

**y – y _{1} = m(x – x_{1})**

**3****. Two Point Form:**

The equation of the line passing through two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is

**4.Intercept-Intercept Form:**

If a line makes intercepts a and b on x and y axis respectively, then its equation is given by:

**Example 6:** ABCD is a square with A (3, 2) and C (7,-4). Find the equation of the diagonal BD.

**Answer: **

The diagonals of a square bisect each other and are perpendicular to each other. So, O becomes the midpoint of AC. Thus, the coordinates of O are

**Example 7:** ABC is a triangle where the equation of AB is 2x –y = 3, BC is x-2y+1 =0 and AC is y+1 = 0. What is the equation of the altitude drawn from A on BC?

**Solution :**

A is the point of intersection of the lines 2x – y – 3 = 0 and y +1 = 0. So coordinates of A become (1,-1).

The slope of BC is -1/2

So, slope of altitude drawn from A = 2

Equation of AD becomes

y +1 = 2(x-1)

y+1 = 2x – 2

or 2x –y = 3

**Example 8:**** Find the image of the point (3,2) about the line x-y = 7.**

**Solution: **

We need to find the image of (3,2) about the line x-y =7

Now as B is the image of point A, so AC = BC and also AB is perpendicular to x-y =7

__Distance between two Parallel Lines:__

As slope of a straight line given by ax+by =c is (-a/b), so if two lines lines are parallel then they have the same slopes, which means if the equation of one line is ax +by + c_{1} = 0, then the equation of the second line is given by ax+by +c_{2} = 0

The distance between two parallel lines ax+by+c_{1} = 0 and ax + by + c_{2} = 0 is given by :

The distance of a point (h,k) from a line ax + by + c = 0 is given by :

**Example 9:** The four sides of a quadrilateral are given by the equations x-3y+4=0, 5x+3y-10=0 , x-3y-6=0 and 5x+3y+8=0. What is the area of this quadrilateral?

**Answer: **Since the slopes of 2x-y+3 =0 and 2x-y-2=0 are equal, so these lines are parallel to each other.

Similarly 5x+y-12=0 and 5x+y-5=0 have the same slope and so they are parallel to each other.

Thus, the quadrilateral is a parallelogram.

Length of the base is given by the distance between the points of intersection of x-3y+4 =0 with the line 5x+3y+8=0 and with 5x+3y-10=0.

Thus, Coordinates of A and B become (-2,2/3) and (1,5/3)

**Example 10: **?ABC is equilateral with A (2,1) and equation of BC is given by 4x -3y + 4 =0. What is the area of ?ABC?

**Solution:** Given to us is the vertex A and equation of BC