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Contents

1. INTRODUCTION
2. THE COORDINATE SYSTEM
   1. QUADRANTS, AXES AND ORIGIN
3. DISTANCE FORMULA AND SECTION FORMULA
   1. THE DISTANCE FORMULA
   2. THE SECTION FORMULA
4. LINES
   1. SLOPE OF A LINE
   2. EQUATION OF A LINE
   3. POINT OF INTERSECTION OF TWO LINES
   4. DISTANCE OF A LINE FROM A POINT
   5. DISTANCE BETWEEN TWO PARALLEL LINES
   6. ACUTE ANGLE BETWEEN TWO LINES
5. CHANGE OF ORIGIN
6. TRIANGLES
7. PARALLELOGRAM
8. EQUATIONS OF COMMON CURVES
   1. EQUATION OF A CIRCLE
   2. EQUATION OF AN ELLIPSE

Coordinate Geometry

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INTRODUCTION

Coordinate geometry is a branch of mathematics that uses the principles of algebra to study geometry. Figures like lines and circles can be represented using algebraic equations and their properties can be studied using these equations. Coordinate geometry also helps us understand the behaviour of functions.

A one-dimensional coordinate system, used to represent points, is the number line similar to the one that we have studied in Number Systems.

A two-dimensional coordinate system is used to represent two-dimensional figures like lines, circles and other curves.

THE COORDINATE SYSTEM

Two perpendicular number lines, XOX’ and YOY’, intersecting at O form the coordinate system.

QUADRANTS, AXES AND ORIGIN

The line XOX’ is called the x-axis and the line YOY’ is called the y-axis.

The x-axis and the y-axis are called the coordinate axes. The point O is called the origin. The axes divide the plane on which they are drawn into four parts known as quadrants and the plane is called thex-y plane.

1. COORDINATES OF A POINT

Any point on the x-y plane can be identified by coordinates. The x-coordinate is the distance of the point from the y-axis and the y-coordinate is the distance of the point from the x-axis. The x-coordinate is also known as the abscissa and the y-coordinate is also known as the ordinate. Any point can be represented by an ordered pair of x and y coordinates as (x, y). For example a point A with x-coordinate 4 and y-coordinate 3 can be represented as A (4, 3). The distance of this point A is 4 units from the y-axis and 3 units from the x-axis.

The signs of the x and y-coordinates of a point change depending on the quadrant in which it lies.

Quadrant	x-coordinate	y-coordinate
I	Positive	Positive
II	Negative	Positive
III	Negative	Negative
IV	Positive	Negative

REMEMBER

• The points lying on the axes are not considered to be in any quadrant.

DISTANCE FORMULA AND SECTION FORMULA

THE DISTANCE FORMULA

The distance between any two points lying on a line parallel to the x-axis and having x-coordinates x₁ and x₂ is |x₂ – x₁|.

Similarly, the distance between any two points lying on a line parallel to the y-axis and having y-coordinates y₁ and y₂ is |y₂ – y₁|.

Distance between two points A(x₁, y₁) and B(x₂, y₂) is given by

Explanation:

Consider points A(x₁, y₁), B(x₂, y₂), C(x₁, y₂) as shown in the figure.

BC = (x₁ – x₂) and AC = (y₁ y₂)

Applying Pythagoras theorem in right angled ΔABC,

(AB)² = (BC)² + (AC)²

(AB)² = (x₁ x₂)² + (y₁ y₂)²

Thus,

Example 1:

Find the distance between the points A(3, 4) and B(7, 3). Solution: Using the formula for distance between two points, units

Example 2:

Distance between the points A(2, 3) and B(x, y) is 5 units while the distance between points B(x, y) and C(6, 1) is 13 units. Find the number of possible values for coordinates of B if B lies in the first quadrant and its coordinates are integers. Solution: B lies in the first quadrant. Its coordinates are positive integers. Distance between two points Squaring both sides, (2 – x)2 + (3 – y)2 = 25 x2 + y2 – 4x – 6y = 12…(i) Similarly, (x + 6)2 + (y – 1)2 = 169 x2 + y2 + 12x – 2y = 132…(ii) Subtracting (i) from (ii) we have, 16x + 4y = 120 4x + y = 30 y = 30 – 4x As x and y are positive integers, substituting the values of x from 1 to 7 we have the possible values of (x, y) as: (7, 2), (6, 6), (5, 10), (4, 14),(3, 18), (2, 22), (1, 26). But, since these values must satisfy equation (i) and (ii) we get (6, 6) as the only possible pair satisfying the given conditions. Hence, possible number of values for coordinates of B satisfying the given conditions is 1.

THE SECTION FORMULA

1. Internal Division

Coordinates of a point C(x, y), which divides the line joining the points A(x₁, y₁) and B(x₂, y₂) internally in the ratio a : b are given by

This is known as the section formula for internal division.

2. External Division

Coordinates of a point C(x, y), which divides the line joining the points A(x₁, y₁) and B(x₂, y₂) externally in the ratio a : b is given by

This is known as the section formula for external division.

REMEMBER

• The formula for external division can be derived using the formula for internal division by considering the fact that point A divides segment BC in the ratio (b – a) : a.

Example 3:

Find the coordinates of the point M dividing line segment AB internally in the ratio 3 : 7 if A (1, 6) and B (3, 11). Solution: Let the coordinates of M be (x, y). Applying section formula of internal division, M (1.6, 7.5)

3. The Midpoint Formula

If a : b 1 : 1, C is the midpoint of line AB. Using the section formula, the coordinates of C are:

Example 4:

If A (2, 3) and B (8, 9). Point C externally divides AB in the ratio 2 : 1. Find the coordinates of C. Solution: As point C divides segment AB in the ratio 2 : 1, we can say that B divides AC in the ratio 1 : 1. Thus, B is the mid-point of AC. x = 14 and y = 15 C (14, 15)

LINES

Positive and Negative Angles

The angles measured anti-clockwise from the positive direction of the x-axis are considered to be positive. In the following figure, the first two angles are positive.

The angles measured clockwise from the positive direction of the x-axis are considered to be negative. The third angle is a negative angle.

SLOPE OF A LINE

The slope of a line, generally denoted by m, is the slant of a line. The slope of a horizontal line is 0. Its magnitude goes on increasing as we gradually make the line vertical as shown.

Clearly, the slope depends on the angle that it makes with the horizontal. It can be calculated in two ways.

1. If the angle made by the line with the positive direction of the x-axis is known, and is say , the slope is given by:

m = tan

is called the inclination of the line.

Slopes can be positive or negative depending on . This is illustrated in the following table.

If A(x₁, y₁) and B(x₂, y₂) are any two points on a line, the slope is given by:

Slopes of Parallel Lines

Parallel lines make equal angles with the positive direction of the x-axis. Hence, the slopes of parallel lines are always equal.

Slopes of Perpendicular Lines

Consider two perpendicular lines having inclinations ₁ and ₂, such that ₂ = (₁ + 90). Let their respective slopes be m₁ and m₂.

Then, m₁ = tan ₁ and m₂ = tan ₂

m₂ = tan (₁ + 90) = –cot ₁

m₁ m₂ = tan ₁ (–cot ₁) = –1

REMEMBER

• Slope of the x-axis is 0 and the slope of the y-axis is ∞.

Example 5:

If the slope of a line is 1/2, then find the slope of a line parallel to this line and a line perpendicular to this line. Solution: Slope of two parallel lines is equal. Hence, slope of a line parallel to this line = 1/2. For two perpendicular lines, m1 m2 = 1 Here, m1 = 1/2 1/2 m2 = 1 m2 = 2 Hence, slope of a line perpendicular to the given line is 2.

EQUATION OF A LINE

As stated earlier, coordinate geometry helps us study geometry using algebraic principles. We can describe a line completely using an algebraic equation. Now let us first try to understand how a line is described.

We have seen the concept of slope. It denotes the slant of a line. If we consider a line with slope 1, we know that it makes an angle of tan^–1(1) = 45 with the positive direction of the x-axis.

All the above lines have slope 1. But all the lines are different. If we know any one point on the line in addition to the slope, we will know exactly where the line lies. Thus, to describe a line completely, we need

1. A point on the line

2. Its slope

Using these two parameters, we can form the equation of a line. The equation of a line can be stated in various forms as described below.

1. SLOPE-POINT FORM

The equation of a line having slope m and passing through the point P(x₁, y₁) is (y – y₁) = m(x – x₁).

Explanation:

If Q(x, y) is another point on the line apart from P(x₁, y₁), the slope of the line is

But the slope is m.

(y – y₁) = m(x – x₁)

Example 6:

Solution: The slope-point form of the equation of a line is (y – y1) = m(x – x1) The equation of the given line is 4y – 12 = 3x – 6 3x – 4y + 6 = 0

REMEMBER

• Any point on a line always satisfies the equation of the line.

2. TWO POINT FORM

As stated earlier, we need a point on a line and the slope to write the equation of the line. When we have two points, we can find the slope. Using this slope and any one of the given points, the equation of the line can be found.

The equation of a line passing through points P(x₁, y₁) and Q(x₂, y₂), when x₁ ≠ x₂, is given by

Explanation:

Using the slope-point form,

(y – y₁) = m(x – x₁)

Using the definition of slope,

Example 7:

Find the equation of the line, which passes through the points (3, 4) and (1, 2). Solution: Two-point form of the equation of a line is This line passes through (3, 4) and (1, 2). Thus, y 4 = x 3 y x 1 = 0 This is the equation of the required line. The equation of the line can be verified by substituting points (3, 4) and (1, 2) in the Both these points lie on the line and hence, must satisfy the equation of the line.

3. SLOPE-INTERCEPT FORM

The equation of a line having slope m and making an intercept c on the y-axis is given by

y = mx + c

Explanation:

Let P(x, y) be any point on the line. Let the line meet the y-axis at point Q. As the y-intercept of the line is c, Q (0, c).

Using the slope-point form,

(y – y₁) = m(x – x₁)

y = mx + c

Example 8:

In the above figure, the angle made by the line AB with the positive direction of the x-axis is 135 and OB = 3. Find equation of the line AB. Solution: Using the slope-intercept form, y = mx + c m = tan = tan 135= –tan 45 = –1 c = OB = 3 Hence, equation of the line is y = –x + 3

The general form of the equation of a line is given as Ax + By + C = 0.

This can also be written as By = Ax C

Hence,

Comparing with the slope-intercept form y = mx + c, we get,

Example 9:

Find the equation of a line parallel to 3y – 4x + 7 = 0 and passing through the point (1, 2). Solution: 3y 4x + 7 = 0 can be written as 4x + 3y + 7 = 0. Comparing with Ax + By + C = 0, A = 4, B = 3 and C = 7 Slope of the required line can be given as: We are given a point (1, 2) lying on this line. So, using the slope point form, (y y1) = m(x x1) 3y 6 = 4x – 4 4x 3y + 2 = 0 This is the equation of the required line.

Example 10:

Find the slope of a line perpendicular to 2y = 4x + 3. Solution: 2y = 4x + 3 4x 2y + 3 = 0 Comparing with Ax + By + C = 0, we get A = 4, B = 2, C = 3 Slope (m) = A/B = 2 The product of the slope of a line and that of a line perpendicular to it is 1.

Example 11:

A line passes through points (1, 3) and (7, 9). Find the equation of a line perpendicular to this line and passing through (2, 5). Solution: Slope of the line passing through (1, 3) and (7, 9), Slope of a line perpendicular to the given line is 1. Also, we know that point (2, 5) lies on this line. So, we now use the slope-point form, (y y1) = m(x x1) Here, x1 = 2, y1 = 5 and m = 1. (y 5) = 1(x 2) y 5 = x + 2 x + y = 7 This is the equation of the required line.

4. TWO-INTERCEPT FORM

The equation of a line forming intercepts a and b (a ≠ 0 and b ≠ 0) on the x-axis and y-axis respectively, is

Explanation:

The line forms intercepts a and b on the x-axis and y-axis respectively.

(a, 0) and (0, b) are two points on the line.

Using the two points form of the equation of a line,

ay + bx = ab

Example 12:

Find the perimeter of a triangle formed by the x-axis, y-axis and the line 3x + 4y = 12. Solution: 3x + 4y = 12 The x-intercept of the line is 4 and the y-intercept is 3. The triangle can be constructed as shown. OB = 4 and OA = 3 Perimeter of the triangle = 3 + 4 + 5 = 12 units

Example 13:

Consider a triangle drawn on the x - y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (x, y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is [CAT 2005] (1) 780(2) 800 (3) 820(4) 741 Solution: The points satisfying the equations x + y < 41, y > 0, x > 0 lie inside the triangle. Integer solutions of x + y < 41 can be found as follows: If x + y = 40, then (x, y) could be (1, 39), (2, 38), …, (39, 1)  ...(39 solutions)  If x + y = 39, then (x, y) could be (1, 38), (2, 37), …, (38, 1)   ...(38 solutions) If x + y = 38, we get 37 solutions and so on till x + y = 2  ... (1 solution) Thus, there are 39 40/2 = 780 integer solutions to x + y < 41. The number of points with integer coordinates lying inside the triangle = 780 Hence, option 1.

REMEMBER

• The equation of the x-axis is y = 0 and the equation of the y-axis is x = 0.

POINT OF INTERSECTION OF TWO LINES

The point, at which two lines intersect, can be found by solving the equations of the two lines simultaneously. The solution satisfies the equations of both the lines and is hence the point of intersection of the two lines.

Example 14:

At what point do the lines 2x + 3y = 5 and x + y = 2 intersect? Solution: 2x + 3y = 5… (i) x+ y = 2 … (ii) Multiplying (ii) by 2 we get 2x + 2y = 4 … (iii) Subtracting (iii) from (i) we get y = 1 Substituting y = 1 in (ii) we get x + 1 = 2 Hence, x = 1 Thus the two lines intersect when x = 1 and y = 1, i.e. at (1, 1).

REMEMBER

• Three or more lines are said to be concurrent lines when all of them pass through a common point.

• If three points A, B, C are collinear (A B C) then any one of the following conditions must be true.

1. Area of triangle ABC = 0

2. Slope of AB = slope of BC = slope of AC

3. AB + BC = AC

DISTANCE OF A LINE FROM A POINT

The distance of a point P(x₁, y₁) from the line Ax + By + C = 0 is

Example 15:

Find the distance of a point (1, 2) from the line 2x + 4y + 3 = 0. Solution: Comparing the given equation with Ax + By + C = 0 we get A = 2, B = 4, C = 3 P (1, 2) x1 = 1, y1 = 2 The distance of the point (1, 2) from the given line can be given as

Example 16:

An isosceles triangle has vertices of the unequal side at (3, 5) and (–5, –1). The area of the triangle is 12.5 sq. units. Find the third vertex of the triangle. Solution: Let ΔABC be the isosceles triangle with AB = AC. Let B (3, 5) and C (–5, –1). As area of the triangle is 12.5 sq. units, AD = 2.5 units As ΔABC is an isosceles triangle, D is the mid-point of BC. D (–1, 2) Let A (m, n). 3n + 4m = 2…(i) 3x – 4y + 11 = 0 Using the formula for the distance of a point from a line, we have, 12.5 = 3m – 4n + 11 3m – 4n = 1.5…(ii) or 3m – 4n = –23.5…(iii) Solving (i) and (ii) simultaneously, m = 0.5 and n = 0 Solving (i) and (iii) simultaneously, m = –2.5 and n = 4 The third vertex of the triangle could be (0.5, 0) or (–2.5, 4).

Example 17:

ΔABC is a right triangle with B = 90. A (1, 10), B (1, 2) and C (7, 2). If BD AC, find the coordinates of D. Solution: Let D (h, k) 3h – 4k = –5…(i) 4x + 3y = 34 As D lies on this line, 4h + 3k = 34…(ii) Solving (i) and (ii) simultaneously,

DISTANCE BETWEEN TWO PARALLEL LINES

The distance between two parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by

Example 18:

Opposite sides of a regular hexagon lie on the lines Find the area of the hexagon.

Solution:

From the figure, GH is the distance between the two parallel lines and this is twice the height of ΔAOB.

Let s be the side of the hexagon.

Example 19:

Find the minimum distance between the lines 2x + 3y + 6 = 0 and 2x – 3y + 4 = 0. Solution: Slope of first line = 2/3. Slope of second line = 2/3. The two lines are not parallel they will intersect and thus the minimum distance between them is 0.

Example 20:

to the given line and 4 units from it. A possible equation for L is: [FMS 2010] Solution: The distance between (0, 6) and L should be 4 units. Distance of a line Ax + By + C = 0 from a point (x1, y1) Consider option 1. Distance of the given line from the point (0, 6) Option 1 is correct. Hence, option 1.

ACUTE ANGLE BETWEEN TWO LINES

Let be the acute angle between two lines with slopes m₁ and m₂ respectively (m₁ m₂ ≠ –1).

Example 21:

Find the acute angle between lines x + y = 0 and x  y = 0. Solution: Slope of first line = 1 Slope of second line = 1 As the product of the slopes of the two lines is –1, the lines are perpendicular to each other.

Example 22:

ΔABC is an isosceles right angled triangle with B = 90, A (0, 3) and C (–2, 7). Find the coordinates of B if it lies in the first quadrant. Solution: mACB = 45 Let slope of BC be m1. Using the formula for the angle between two lines we have, Let the slope of AB be m2. Now we can find the equations of lines BC and AB using the slope-point form. Equation of BC is x + 3y = 19 or y = 3x + 13 Equation of AB is y = 3x + 3 or x + 3y = 9 Solving simultaneously, B (1, 6) or B (–3, 4) But as B lies in the first quadrant, B (1, 6).

CHANGE OF ORIGIN

The origin of the coordinate system is (0, 0). If we wish to shift the origin to another location having coordinates (h, k), axes remaining parallel, then the coordinates of a point P(x, y) become (x – h, y – k) with respect to the new origin.

Example 23:

On shifting the origin to a point P, axes remaining parallel, the coordinates of a point (5, 9) change to (1, –8). Find the coordinates of P. Solution: Let P (h, k) By using the concept of change of origin, 1 = 5 – h and –8 = 9 – k h = 4 and k = 17 P (4, 17)

Example 24:

The equation of a curve is x2 + y2 = 25. If the origin is shifted to point (5, 3), axes remaining parallel, find the new equation of the curve. Solution: Let (x, y) change to (X, Y) when the origin is shifted. Then, X = x – 5 and Y = y – 3 x = X + 5 and y = Y + 3 To find the new equation of the curve in the new system, we replace x and y in the original equation with (X + 5) and (Y + 3) respectively. (X + 5)2 + (Y + 3)2 = 25 X2 + Y2 + 10X + 6Y + 25 + 9 = 25 X2 + Y2 + 10X + 6Y + 9 = 0 Replacing X and Y with x and y, x2 + y2 + 10x + 6y + 9 = 0 is the equation of the curve in the new system.

TRIANGLES

The Centroid

The point of intersection of the three medians of a triangle is its centroid.

In the following figure, G is the centroid of ΔABC. It divides each median internally in the ratio 2:1.

If A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃), the coordinates of the centroid G can be found using the section formula and are given by

The Incentre

The incentre of a triangle is the point of intersection of the angle bisectors of its vertices. It is the centre of a circle that has all the three sides of the triangle tangent to it.

The coordinates of incentre (x_i, y_i) are given by

The Area of a Triangle

The area of triangle with coordinates of vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is given by

This can be proved using distance formula and then applying Heron’s formula.

Example 25:

Find the area of the triangle with A (2, 3), B (4, 5) and C (6, 9) as its vertices. Solution: A = 2 square units Alternatively, If BC is the base, we can find the length of BC using the distance formula. The height of ΔABC, h, is the distance of A from BC. This can be found using the formula for the distance of a point from a line. The equation of BC can be found using the two-point form. The equation of BC is 2x y = 3 Area of ΔABC = 2 square units

Example 26:

The area of the triangle with the vertices (a, a),(a + 1, a) and (a, a + 2) is [CAT 2002] (1) a3(2) 1 (3) 0 (4) None of these Solution: Area of (Δ) = where x1 = y1 = a, x2 = a + 1, y2 = a, x3 = a and y3 = a + 2 Hence, option 2.

PARALLELOGRAM

If a quadrilateral has vertices P(x₁, y₁), Q(x₂, y₂), R(x₃, y₃) and S(x₄, y₄), and x₁ + x₃ = x₂ + x₄ and y₁ + y₃ = y₂ + y₄, the quadrilateral PQRS is a parallelogram. The converse of this statement is also true.

Example 27:

ABCD is a rhombus with the diagonals AC and BD intersecting at the origin on the x-y plane. The equation of the straight line AD is x + y = 1. What is the equation of BC? [CAT 2000] (1) x + y = –1(2) x – y = –1 (3) x + y = 1(4) None of the above Solution: Equation of line AD is x + y = 1. AD intersects x-axis at (1, 0) and y-axis at (0, 1). ABCD is a rhombus, with diagonals AC and BD intersecting at origin. The four vertices are (1, 0), (0, 1), (1, 0) and (0, 1). Since AD passes through (1, 0) and (0, 1), BC passes through (1, 0) and (0, 1). i.e. x + y = 1 Hence, option 1.

EQUATIONS OF COMMON CURVES

EQUATION OF A CIRCLE

Equation of a circle having its centre at (0, 0) and radius of a units is

If the circle has its centre at (h, k) and its radius is a units, its equation is

The general form of the equation of a circle is

x² + y² + 2gx + 2fy + c = 0,

where g² + f² – c > 0

The centre of this circle is (–g, –f) and radius is

Example 28:

What is the length of the tangent drawn from the point (2, −1) to the circle 3x2 + 3y2 + 4x + 2y + 6 = 0? [FMS 2009] (1) 3(2) 6 (3) 9(4) None of these Solution: The standard equation a circle having centre (a, b) and radius r is given by the following equation: (x a)2 + (y b)2 = r2 The equation given in this case is 3x2 + 3y2 + 4x + 2y + 6 = 0 To put it in the standard form, we divide the entire equation by 3. Hence we get, The radius of the circle is an imaginary number, the circle is not real. The length of the tangent from the point (2, 1) cannot be determined. Hence, option 4.

Example 29:

The area of a circle whose centre is at (2, 0) is . The circle does not pass through the following point: [JMET 2009] (1) (2, 1)(2) (2, 1) (3) (3, 0)(4) (2, 2) Solution: The circle whose centre is at (2, 0) and has area , will have a radius of 1 unit. The equation of the circle with centre (a, b) and radius r is (x a)2 + (y b)2 = r2 Equation of the given circle is (x 2)2 + (y 0)2 = 12 On substituting the points given in the options in the equation of the circle, we get that (2, 2) does not satisfy the equation. Hence, option 4.

EQUATION OF AN ELLIPSE

Equation of a standard ellipse having its centre at (0, 0) is given by

where the lengths of its major and minor axes are 2a and 2b respectively.

The area of an ellipse is ab and its perimeter is (a + b)

Example 30:

Find the equation of circle having its centre at origin and radius units. Solution: Equation of the circle is

Example 31:

The graph of x2 – 4y2 = 0 is: [FMS 2010] (1) a parabola (2) an ellipse (3) a pair of straight lines (4) none of these Solution: x2 – 4y2 = 0 (x + 2y) (x – 2y) = 0 x = 2y Thus, the given equation is the equation of a pair of straight lines. Hence, option 3.

REMEMBER

• Distance between two points A(x₁, y₁) and B(x₂, y₂) is given by

• Coordinates of a point C(x, y), which divides the line joining the points A(x₁, y₁) and B(x₂, y₂) internally in the ratio a : b are given by

• Coordinates of a point C(x, y), which divides the line joining the points A(x₁, y₁) and B(x₂, y₂) externally in the ratio a : b is given by

• Slope of a line = tan , where is the inclination of the line.

• If A(x₁, y₁) and B(x₂, y₂) are any two points on a line, the slope is given by:

• Slopes of parallel lines are always equal.

• Product of the slopes of two perpendicular lines is –1.

• The equation of a line having slope m and passing through the point P(x₁, y₁) is:

(y – y₁) = m(x – x₁).

• The equation of a line passing through points P(x₁, y₁) and Q(x₂, y₂), when x₁ ≠ x₂, is given by

• The equation of a line having slope m and making an intercept c on the y-axis is given by:

y = mx + c.