Coordinate geometry: Every planet in the Universe has coordinates, which help us locate them easily. The Earth has two imaginary lines in its coordinate system, known as longitudes and latitudes. These coordinates help us to locate any place on Earth.

Similarly, we can locate the point on the paper or plane by using coordinate axes such as the horizontal axis (\(x-\)axis) and vertical axis (\(y-\)axis).

Introduction to Coordinate Geometry:

Coordinate geometry is a particular branch of mathematics that gives the relation between algebra and geometry by using graphs. Coordinate geometry helps us find the point’s position on the plane by using the ordered pairs.

Two imaginary lines mainly constitute coordinate geometry: one is a vertical line, and the other is horizontal. These two lines are perpendicular to each other. The horizontal line is called the \(x-\)axis, and the vertical line is called the \(y-\)axis.

The two axes (\(x-\)axis and \(y-\)axis) intersect at one point perpendicularly. The intersection point of the \(y-\)axis and \(x-\)axis is called the origin, and its coordinates are given by \((0, 0.)\)

Coordinate Plane

As we know, the coordinate plane is divided into four parts with two imaginary lines, which are perpendicular to each other. The position of any point can be located on the plane or paper by using the coordinate plane.

The points are located in the coordinate plane or cartesian plane in the ordered pairs \((x, y.)\)

The values of the ordered pair are known as coordinates of the point. The first value of the point is called the abscissa, and the second value is called the ordinate of the point.

The cartesian plane is divided by two axes (\(x-\)axis and \(y-\)axis) into four parts known as quadrants. 

1. Quadrant-1:
In this quadrant, both coordinates of the point are positive.
\(P (+x, +y)\)

2. Quadrant-2:
In this quadrant, \(x-\)coordinate is negative, and \(y-\)coordinate is positive.
\(P (-x, +y)\)

3. Quadrant-3:
In this quadrant, \(x-\)coordinate is negative, and \(y-\)coordinate negative.
\(P (-x, -y)\)

4. Quadrant-4:
In this quadrant, \(x-\)coordinate is positive, and \(y-\)coordinate is negative.
\(P (+x, -y)\)

Important points to remember:

1. The \(x-\)axis to the right of origin is positive \(x-\)axis.
2. The \(x-\)axis to the left of the origin is the negative \(x-\)axis.
3. The \(y-\)axis, which is above the origin, is a positive \(y-\)axis.
4. The \(y-\)axis, which is below the origin, is the negative \(y-\)axis.
5. The intersection point of \(y-\)axis and \(x-\)axis is origin \(O (0, 0.)\)
6. The point, whose abscissa is zero, lies on \(y-\)axis.
7. The point whose ordinate is zero lies on \(x-\)axis.

Distance Formula in Coordinate Geometry

Consider two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) on the cartesian plane.

From the above figure,

\(BC = {x_2} – {x_1}\) and \(AC = {y_2} – {y_1}\)

In the right triangle \(ABC,\) by Pythagoras theorem,

\( \Rightarrow A{B^2} = B{C^2} + A{C^2}\)

\( \Rightarrow A{B^2} = {\left( {{x_2} – {x_1}} \right)^2} + {\left( {{y_2} – {y_1}} \right)^2}\)

\( \Rightarrow AB = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \)

Therefore, the distance between two points \(\left( {{x_1},{y_1}} \right)\) and \(\left( {{x_2},{y_2}} \right)\) is given by the formula:

\(\sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \)

Section Formula in Coordinate Geometry:

Consider two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) on the cartesian plane and the point \(P (x, y)\) in the ratio \(m:n\) as shown below:

In two right triangles \(APQ,\) and \(BPC\)

\(\angle PAQ = \angle BPC\) ( \(∵\) Corresponding angles )

\(\angle AQP = \angle PCB = {90^{\rm{o}}}\)

By \(A.A\) similarity criterion, \(\Delta PAQ \sim \Delta BPC\)

So, by CPCT 

\( \Rightarrow \frac{{AP}}{{PB}} = \frac{{AQ}}{{PC}}\) and \(\frac{{AP}}{{PB}} = \frac{{PQ}}{{BC}}\)

\( \Rightarrow \frac{m}{n} = \frac{{x – {x_1}}}{{{x_2} – x}}\) and \(\frac{m}{n} = \frac{{y – {y_1}}}{{{y_2} – y}}\)

\( \Rightarrow m\left( {{x_2} – x} \right) = n\left( {x – {x_1}} \right)\) and \(m\left( {{y_2} – y} \right) = n\left( {y – {y_1}} \right)\)

\( \Rightarrow m{x_2} – mx = nx – n{x_1}\) and \(m{y_2} – my = ny – n{y_1}\)

\( \Rightarrow m{x_2} + n{x_1} = mx + nx\) and \(m{y_2} + n{y_1} = ny + my\)

\( \Rightarrow x(m + n) = m{x_2} + n{x_1}\) and \(y(m + n) = m{y_2} + n{y_1}\)

\( \Rightarrow x = \frac{{m{x_2} + n{x_1}}}{{m + n}}\) and \(y = \frac{{m{y_2} + n{y_1}}}{{m + n}}\)

Therefore, coordinates of the point \(P,\) which divides the line segment joining the points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) in the ratio \(m:n\) internally is given by

\(P(x,y) = \left[ {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right]\)

Mid-Point Formula in Coordinate Geometry

Let us consider a point \(P (x, y),\) which is the mid-point of the line segment joining two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right).\)

Mid-point divides the given line in the ratio of \(1:1.\)

So, \(m:n=1:1\)

By section formula, the coordinates of the point \(P,\) which divides the line segment joining the points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) in the ratio \(m:n\) internally is given by:

\(P(x,y) = \left[ {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right]\)

\( \Rightarrow P(x,y) = \left[ {\frac{{(1){x_2} + (1){x_1}}}{{1 + 1}},\frac{{(1){y_2} + (1){y_1}}}{{1 + 1}}} \right]\)

 \( \Rightarrow P(x,y) = \left[ {\frac{{{x_2} + {x_1}}}{2},\frac{{{y_2} + {y_1}}}{2}} \right]\)

Therefore, the coordinates of the point \(P,\) which is the mid-point of the line segment joining the two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) is given by:

\(P\left({x,y} \right) = \left[{\frac{{{x_2} + {x_1}}}{2},\frac{{{y_2} + {y_1}}}{2}} \right]\)

The slope of a Line in Coordinate Plane:

Consider the line segment joining two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\)

From the below figure,

\(BC = {x_2} – {x_1}\) and \(AC = {y_2} – {y_1}\)

The tangent of the angle gives the slope of the line. 

In the triangle \(ABC,\) \(\tan \,tan\,\theta = \frac{{AC}}{{BC}}\)

\( \Rightarrow \tan tan\,\theta = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

Generally, the slope of the line is represented by \(“m”.\)

Therefore, the slope of the line passing through two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) is given by

\(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

Area of Triangle in Coordinate Geometry

We know that triangle is a three-dimensional polygon in geometry. Space occupied by the triangle in the given two-dimensional plane is known as the area of the triangle.

Consider a triangle with three vertices \(A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)\) and \(C\left( {{x_3},{y_3}} \right).\)

The area of triangle formula in coordinate geometry can be used in the following steps:

1. Plot the given three points \(A\left( {{x_1},\,{y_1}} \right),B\left( {{x_2},\,{y_2}} \right)\) and \(C\left( {{x_3},\,{y_3}} \right)\) on the coordinate plane.

2. Consider all the values in the counter-clockwise direction only.

3. Join the points \(AB, BC\) and \(AC\) to form a triangle \(ABC\) in the coordinate plane.

4. Arrange the given vertices in the following manner:

5. Find the product of diagonal elements as shown with a blue colour arrow and add the products obtained.
\({x_1}{y_2} + {x_2}{y_3} + {x_2}{y_1} \ldots (i)\)

6. Find the product of diagonal elements as shown with a red colour dotted arrow and add the products.
\({x_2}{y_1} + {x_3}{y_2} + {x_1}{y_3} \ldots (ii)\)

7. Now, subtract the equation \({\rm{(ii)}}\) from equation \({\rm{(i)}},\) which is multiplied with \(\frac{1}{2}\) to give the area of the triangle.
 \(\frac{1}{2}\left[ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_1}} \right) – \left( {{x_2}{y_1} + {x_3}{y_2} + {x_1}{y_3}} \right)} \right]\)

8. After simplifying the above formula, we can write
 \(\frac{1}{2}\left[ {{x_1}\left( {{y_2} – {y_3}} \right) + {x_2}\left( {{y_3} – {y_1}} \right) + {x_3}\left( {{y_1} – {y_2}} \right)} \right]\)

Therefore, the area of the triangle \(ABC,\) with three vertices \(A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)\) and \(C\left( {{x_3},{y_3}} \right)\) is given by

\(\frac{1}{2}\left[ {{x_1}\left( {{y_2} – {y_3}} \right) + {x_2}\left( {{y_3} – {y_1}} \right) + {x_3}\left( {{y_1} – {y_2}} \right)} \right]\)

Applications of Coordinate Geometry

Some of the applications of coordinate geometry are 

1. To locate the position of cursors in computers, mobiles etc.
2. In maps for navigation (GPS).
3. To find geographical locations.
4. It finds its application in some mathematical problems like algebra, calculus, functions, complex numbers etc.

Solved Examples – Coordinate Geometry:

The below problems will help in understanding the coordinate geometry formulas.

Q.1. Find the area of the triangle formed with three vertices that are \((6, 7,) (2, -9)\) and \((-4, 1.)\)
Ans: Let three vertices of the triangle \(ABC\) be \(A (6, 7), B (2, -9)\) and \(C (-4, 1).\)
We know that the area of the triangle \(ABC,\) with three vertices \(A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)\) and \(C\left( {{x_3},{y_3}} \right)\) is given by \(\frac{1}{2}\left[ {{x_1}\left( {{y_2} – {y_3}} \right) + {x_2}\left( {{y_3} – {y_1}} \right) + {x_3}\left( {{y_1} – {y_2}} \right)} \right].\)
\( = \frac{1}{2}[6(( – 9) – 1) + 2(1 – 7) + ( – 4)(7 – ( – 9))]\)
\( = \frac{1}{2}[6( – 10) + 2( – 6) – 4(16)]\)
\( = \frac{1}{2}[ – 60 – 12 – 64]\)
\( = \frac{1}{2}[ – 136]\)
\( = – 68\)
We know that area cannot be negative, so the area is \({\rm{68}}\,{\rm{sq}}{\rm{.units}}{\rm{.}}\)
Hence, the area of the triangle with given coordinates is \({\rm{68}}\,{\rm{sq}}{\rm{.units}}{\rm{.}}\)

Q.2. The embibe colony is mapped on the coordinate plane. The house of Asit is located at the point \((8, -6),\) and the house of Preethu is located at the point \((-4, 6.)\) And, the house of Keerthi is located halfway between these two houses. Find the coordinates of the point, which represents the house of Keerthi.
Ans: Given the house of the Asit is located at the point \((8, -6),\) and the house of the Preethu is located at the point \((-4, 6.)\)
And, the house of Keerthi is located halfway between the houses of Asit and Preethu.
Thus, Keerthi’s house is at the midpoint between Asit and Preethu’s houses.

We know that the coordinates of the point \(P,\) which is the mid-point of the line segment joining the two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) is given by \(P(x,y) = \left[ {\frac{{{x_2} + {x_1}}}{2},\frac{{{y_2} + {y_1}}}{2}} \right].\)
\( \Rightarrow (x,y) = \left[ {\frac{{8 + ( – 4)}}{2},\frac{{6 + ( – 6)}}{2}} \right]\)
\( \Rightarrow (x,y) = \left[ {\frac{4}{2},\frac{0}{2}} \right]\)
\( \Rightarrow (x,y) = (2,0)\)
Therefore Keerthi’s house is located at the point \((2, 0.)\)

Q.3. Find the slope of the line passing through the points \((2, 3)\) and \((4, 6).\)
Ans: Let the points are \(A (2, 3)\) and \(B (4, 6).\)
We know that the slope of the line passing through two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) is given by \(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}.\)
\( \Rightarrow m = \frac{{6 – 3}}{{4 – 2}}\)
\( \Rightarrow m = \frac{3}{2}\)
Hence, the slope of the line passing through the points \((2, 3)\) and \((4, 6)\) is \(\frac{3}{2}.\)

Q.4. Cherry drew two points, \(A (4, 5\)) and \(B (-3, 8)\) on the coordinate plane. Find the distance between the points drawn by Cherry on the coordinate plane.
Ans: Given the points drawn by Cherry are \(A (4, 5)\) and \(B (-3, 8).\)
We know that the distance between two points \(\left( {{x_1},{y_1}} \right)\) and \(\left( {{x_2},{y_2}} \right)\) is given by \(\sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \)
\( \Rightarrow AB = \sqrt {{{( – 3 – 4)}^2} + {{(8 – 5)}^2}} \)
\( \Rightarrow AB = \sqrt {{{( – 7)}^2} + {3^2}} \)
\( \Rightarrow AB = \sqrt {49 + 9} \)
\( \Rightarrow AB = \sqrt {58} \)
Hence, the distance between the points drawn by the Cherry is \(\sqrt {58} \,{\rm{units}}{\rm{.}}\)

Q.5. Find the ratio in which the point \(P (-7, -3)\) divides the line segment joining the points \(A(-3, -1)\) and \(B (-1, 0).\)
Ans: The points forming the line segment are \(A(-3, -1)\) and \(B (-1, 0).\)
Let the ratio in which the point \(P (-7, -3)\) is \(m:n.\)
We know that the coordinates of the point \(P,\) which divides the line segment joining the points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) in the ratio m:n internally is given by \(P(x,y) = \left[ {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + m{y_1}}}{{m + n}}} \right].\)
\( \Rightarrow ( – 7, – 3) = \left[ {\frac{{m( – 1) + n( – 3)}}{{m + n}},\frac{{m(0) + n( – 1)}}{{m + n}}} \right]\)
Equating the \(x-\)coordinates of the given points,
\( \Rightarrow – 7 = \frac{{ – m – 3u}}{{m + n}}\)
\( \Rightarrow – 7(m + n) = – m – 3n\)
\( \Rightarrow – 7m – 7n = – m – 3n\)
\( \Rightarrow – 7n + 3n = – m + 7m\)
\( \Rightarrow – 4n = 6m\)
\( \Rightarrow \frac{m}{n} = – \frac{6}{4} = – \frac{2}{3}\)
Thus, the ratio \(m:n=-2:3\) or \(2:3\) externally.
Therefore, the ratio in which the point \(P (-7, -3)\) divides the line segment joining the points \(A(-3, -1)\) and \(B (-1, 0)\) is \(2:3\) externally or \(-2:3.\)

Summary:

In this article, we have studied the definition of coordinate geometry. We have learned the meanings of coordinates, axes, origin and the coordinate plane, and also the quadrants formed by the axes. 

Here we have studied the different concepts used and essential points related to coordinate geometry. We also looked at the slope of the line passing through two points in the coordinate plane.

We have studied the formulas of coordinate geometry such as distance formula, section formula, midpoint formula and the formula of area of the triangle.

Frequently Asked Questions (FAQ) – Coordinate Geometry:

Q.1. What is coordinate geometry?
Ans: Coordinate geometry gives the relationship between algebra and geometry by using graphs. It helps us find the point’s position on the plane by using the ordered pairs.

Q.2. How to find the ratio in coordinate geometry?
Ans: The ratio can be found by using the section formula, the coordinates of the point \(P,\) which divides the line segment joining the points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) in the ratio m:n internally is given by \(P(x,y) = \left[ {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right].\)

Q.3. How to find the area of the triangle in coordinate geometry?
Ans: The area of the triangle \(ABC,\) with three vertices \(A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)\) and \(C\left( {{x_3},{y_3}} \right)\) is given by \(\frac{1}{2}\left[ {{x_1}\left( {{y_2} – {y_3}} \right) + {x_2}\left( {{y_3} – {y_1}} \right) + {x_3}\left( {{y_1} – {y_2}} \right)} \right].\)

Q.4. What is the distance formula in coordinate geometry?
Ans: The distance between two points \(\left( {{x_1},\,{y_1}} \right)\) and \(\left( {{x_2},\,{y_2}} \right)\) is given by \(\sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} .\)

Q.5. What is the slope in the coordinate geometry?
Ans: The slope of the line passing through two points \(A\left( {{x_1},\,{y_1}} \right)\) and \(B\left( {{x_2},\,{y_2}} \right)\) is given by \(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}.\)