**CONTENT LIST**

## Coordinate geometry

**Coordinate geometry is the topic in mathematics which deals coordinates
of a point, coordinate axes, Cartesian system, plotting of a point, etc. **

Coordinates in Cartesian plane:

**The Cartesian plane is a two-dimensional
coordinate plane, which is formed by the intersection of two perpendicular
lines. The horizontal line is called as X-axis or abscissa, and the vertical line
is called as Y-axis or ordinates.**

**The point at which the x-axis and y-axis intersect
is called the origin. In the Cartesian plane, the origin is represented by an
ordered pair or the Cartesian coordinate (0, 0).**

**Any coordinate point (x, y) on the Cartesian
plane resents a point, that have horizontal distance of the point from the
origin is x, and the vertical distance is y.**

** **

**The points on the right side and above of the
origin is taken as positive. **

** **

**The points on the right side and below of the
origin is taken as negative. **

** **

**The points on the left side and above of the
origin is taken as positive. **

**The points on the left side and below of the
origin is taken as negative. **

## How to locate the points in the Cartesian plane?

**To locate the point (2, 5) in the Cartesian plane, **

**Find the abscissa (x-value) and ordinate (y-value) from the given ordered pair. Here we get, Abscissa = 2 and ordinate= 5.****Next, plot the value of x =“2” unit on the x-axis. We move 2 unit from the origin to the right on the x axis, as given value of x is positive.****Next, plot the value of y =“5” unit on the y axis. We move 5 units up parallel to the y-axis, as given value of y the positive.**

**Finally, the point A (2, 5) is located on the first quadrant of the
Cartesian plane. **

## Distance between two points in a coordinate system** **

**The distance formula is a formula which help to determine the distance between two points in a coordinate system.**

**Distance**

**between point A and B= c**

**(AB)**

^{2}= √[(BC)^{2}+ (CA)^{2}]**c = √[(x**

_{2}– x_{1})^{2}+ (y_{2}– y_{1})^{2}]**Midpoint coordinates: **

**Midpoint coordinates (x,y) of any two points (x₁,y₁) and (x₂,y₂), is determined by **

** x = ****(x₁ + x₂)/2**

**y = ****(y₁ + y₂)/2**

**Section Formula:**

**Section formula helps to coordinates of a point, which divides the line segment joining any two given points in a given ratio.**

**Given any two points A(x1, y1) and B(x2, y2). We have to find the coordinates of the point C (X,Y) which divides AB in the ratio m : n, i.e. AC / CB = m / n.**

**CASE 1:-**

**When C lying between A and B, ****coordinate**** of point C(x,y)**

**CASE 2:-**

**When C lying outside A and B, coordinate of point C****(x,y)**

##

Collinear Points

**Collinear points mean points lying in a single line.**

## Collinear Points Formula

**There are three methods to determine the collinear points.**

**Distance Formula**

**Three points A, B and C are said to be collinear if AB + BC = AC **

**Slope FormulaThree points A, B and C are said to be collinear if slop of AB = slop of BC.(y2 - y1)/(x2 - x1) = (y3 - y2)/(x3 - x2)**

**Three points A, B and C are said to be collinear if Area of triangle ABC = 0(1/2) [x(y − y) + x(y − y) + x(y − y)] = 0**

## Problems based on coordinate geometry

**Question 1.**

If the points (a,0), (0,b) and (1,1) are collinear, find the value of 1/a + 1/b.

If the points (a,0), (0,b) and (1,1) are collinear, find the value of 1/a + 1/b.

**Solution :**

**Given, collinear points => (a,0), (0,b) and (1,1)**

So, the slop of line joining (a,0) and (0,b) is equal to the slop of line joining (0,b) and (1,1)

Slop of he line joining (a,0) and (0,b)

=> (b - 0)/(0 - a)

=> -b/a

Slop of he line joining (0,b) and (1,1)

=> (1 - b)/(1 - 0)

=> 1 - b

=> -b/a = 1 - b

=> -1/a = (1 - b)/b

=> -1/a = 1/b - 1

=> 1 = 1/b + 1/a

=> 1/a + 1/b = 1

So, the value of 1/a + 1/b is 1.

So, the slop of line joining (a,0) and (0,b) is equal to the slop of line joining (0,b) and (1,1)

Slop of he line joining (a,0) and (0,b)

=> (b - 0)/(0 - a)

=> -b/a

Slop of he line joining (0,b) and (1,1)

=> (1 - b)/(1 - 0)

=> 1 - b

=> -b/a = 1 - b

=> -1/a = (1 - b)/b

=> -1/a = 1/b - 1

=> 1 = 1/b + 1/a

=> 1/a + 1/b = 1

So, the value of 1/a + 1/b is 1.

**Question 2.**

If the points (a,0), (0,b) and (4,k) are collinear, find the value of k in terms

of a and b.

If the points (a,0), (0,b) and (4,k) are collinear, find the value of k in terms

of a and b.

**Solution :**

**Given, collinear points => (a,0), (0,b) and (4,k)**

So, the slop of line joining (a,0) and (0,b) is equal to the slop of line joining (0,b) and (4,k)

Slop of he line joining (a,0) and (0,b)

=> (b - 0)/(0 - a)

=> -b/a

Slop of he line joining (0,b) and (4,k)

=> (k - b)/(4 - 0)

=> (k - b)/4

=> -b/a = (k - b)/4

=> -4b/a = (k - b)

=> -4b/a + b = k

=> b - 4b/a = k

=> (ab - 4b)/a = k

So, the value of k is (ab - 4b)/a.

So, the slop of line joining (a,0) and (0,b) is equal to the slop of line joining (0,b) and (4,k)

Slop of he line joining (a,0) and (0,b)

=> (b - 0)/(0 - a)

=> -b/a

Slop of he line joining (0,b) and (4,k)

=> (k - b)/(4 - 0)

=> (k - b)/4

=> -b/a = (k - b)/4

=> -4b/a = (k - b)

=> -4b/a + b = k

=> b - 4b/a = k

=> (ab - 4b)/a = k

So, the value of k is (ab - 4b)/a.

**Question 3.**

If the points A(4,3) and B (x,5) are on the circle with centre O(2,3), then find the value of x.

If the points A(4,3) and B (x,5) are on the circle with centre O(2,3), then find the value of x.

**Solution :**

**Given, points A(4,3) and B (x,5) and centre O(2,3)**

So, distace between A and O = distace between B and O

=> √(4 - 2)² + (3 - 3)² = √(x - 2)² + (5 - 3)²

=> (2)² + (0)² = (x - 2)² + (2)²

=> (4 + 0) = x² + 4 - 4x + 4

=> 4 = x² - 4x + 8

=> 4 = x² - 4x + 8

=> x² - 4x + 4 = 0

=> x² - 2x - 2x + 4 = 0

=> x(x - 2) -2(x - 2) = 0

=> (x - 2)(x - 2) = 0

So, the value of x is 2.

So, distace between A and O = distace between B and O

=> √(4 - 2)² + (3 - 3)² = √(x - 2)² + (5 - 3)²

=> (2)² + (0)² = (x - 2)² + (2)²

=> (4 + 0) = x² + 4 - 4x + 4

=> 4 = x² - 4x + 8

=> 4 = x² - 4x + 8

=> x² - 4x + 4 = 0

=> x² - 2x - 2x + 4 = 0

=> x(x - 2) -2(x - 2) = 0

=> (x - 2)(x - 2) = 0

So, the value of x is 2.

**Question 4.**

Find the value of x for which the distance between the points P(x,4) and Q(9,10) is 10 unit.

Find the value of x for which the distance between the points P(x,4) and Q(9,10) is 10 unit.

**Solution :**

**Given, point P(x,4) and Q(9,10) and distance between points is 10 units.**

According to the question

=> √(x - 9)² + (4 - 10)² = 10

=> √(x² + 81 - 18x) + (-6)² = 10

=> √x² + 81 - 18x + 36 = 10

=> √x² - 18x + 117 = 10

=> x² - 18x + 117 = 100

=> x² - 18x + 17 = 0

=> x² - x - 17x + 17 = 0

=> x(x - 1) -17(x - 1) = 0

=> (x - 1)(x - 17) = 0

So, the values of x are 1 and 17.

According to the question

=> √(x - 9)² + (4 - 10)² = 10

=> √(x² + 81 - 18x) + (-6)² = 10

=> √x² + 81 - 18x + 36 = 10

=> √x² - 18x + 117 = 10

=> x² - 18x + 117 = 100

=> x² - 18x + 17 = 0

=> x² - x - 17x + 17 = 0

=> x(x - 1) -17(x - 1) = 0

=> (x - 1)(x - 17) = 0

So, the values of x are 1 and 17.

**Question 5.**

If the point A(0,2) is equidistant from the point B (3,p) and C(p,5), find the value of p. Also, find the length of AB

If the point A(0,2) is equidistant from the point B (3,p) and C(p,5), find the value of p. Also, find the length of AB

**Solution :**

**Given, point A(0,2) is equidistant from the point B (3,p) and C(p,5)**

According to the question

=> √(0 - 3)² + (2 - p)² = √(0 - p)² + (2 - 5)²

=> (-3)² + (2 - p)² = (0 - p)² + (-3)²

=> (2 - p)² = (0 - p)²

=> 4 + p² + 4p = p²

=> 4p = -4

=> p = -1

Putting The value of p for finding the length of AB, we get

AB = √(-3)² + (2 - p)²

AB = √(9 + 4 + p² + 4p)

AB = √(p² + 4p + 13)

AB = √((-1)² + 4(-1) + 13)

AB = √(1 - 4 + 13)

AB = √10

So, the values of p is -1 and the length of AB is √10 units.

According to the question

=> √(0 - 3)² + (2 - p)² = √(0 - p)² + (2 - 5)²

=> (-3)² + (2 - p)² = (0 - p)² + (-3)²

=> (2 - p)² = (0 - p)²

=> 4 + p² + 4p = p²

=> 4p = -4

=> p = -1

Putting The value of p for finding the length of AB, we get

AB = √(-3)² + (2 - p)²

AB = √(9 + 4 + p² + 4p)

AB = √(p² + 4p + 13)

AB = √((-1)² + 4(-1) + 13)

AB = √(1 - 4 + 13)

AB = √10

So, the values of p is -1 and the length of AB is √10 units.

**Question 6.**

Find the coordinates of the point on x axis which is at a distance of 10 units from the points (-2,5) and (2,-3).

Find the coordinates of the point on x axis which is at a distance of 10 units from the points (-2,5) and (2,-3).

**Solution :**

**Let the point on x axis be (x, 0) which is at 10 unit distance from points (-2,5) and (2,-3).**

According to the question

=> √(x + 2)² + (0 - 5)² = 10

=> √(x - 2)² + (0 + 3)² = 10

=> √(x + 2)² + (0 - 5)² = √(x - 2)² + (0 + 3)²

=> (x + 2)² + (0 - 5)² = (x - 2)² + (0 + 3)²

=> x² + 4 + 4x + 25 = x² + 4 - 4x + 9

=> 8x = -16

=> x = -2

So, the values of the point on x axis is (-2,0).

According to the question

=> √(x + 2)² + (0 - 5)² = 10

=> √(x - 2)² + (0 + 3)² = 10

=> √(x + 2)² + (0 - 5)² = √(x - 2)² + (0 + 3)²

=> (x + 2)² + (0 - 5)² = (x - 2)² + (0 + 3)²

=> x² + 4 + 4x + 25 = x² + 4 - 4x + 9

=> 8x = -16

=> x = -2

So, the values of the point on x axis is (-2,0).

**Question 7.**

If the point P(x,y) is equidistant from the points A(5,1) and B(-1,5), prove that 3x = 2y.

If the point P(x,y) is equidistant from the points A(5,1) and B(-1,5), prove that 3x = 2y.

**Solution :**

**Given, point P(x,y) is equidistant from the points A(5,1) and B(-1,5)**

To prove, 3x = 2y

Proof:-

According to the question

=> AP = BP => √(x - 5)² + (y - 1)² = √(x + 1)² + (y - 5)²

=> (x - 5)² + (y - 1)² = (x + 1)² + (y - 5)²

=> x² + 25 - 10x + y² + 1 - 2y = x² + 1 + 2x + y² + 25 - 10y

=> -12x = -8y

=> 3x = 2y [Hence proved]

To prove, 3x = 2y

Proof:-

According to the question

=> AP = BP => √(x - 5)² + (y - 1)² = √(x + 1)² + (y - 5)²

=> (x - 5)² + (y - 1)² = (x + 1)² + (y - 5)²

=> x² + 25 - 10x + y² + 1 - 2y = x² + 1 + 2x + y² + 25 - 10y

=> -12x = -8y

=> 3x = 2y [Hence proved]

**Question 8.**

If P(x,y) is a point equidistant from the points A(6,−1) and B(2,3), show that

x − y = 3.

If P(x,y) is a point equidistant from the points A(6,−1) and B(2,3), show that

x − y = 3.

**Solution :**

**Given, point P(x,y) is equidistant from the points A(6,−1) and B(2,3)**

To prove, x − y = 3

Proof:-

According to the question

=> AP = BP => √(x - 6)² + (y + 1)² = √(x - 2)² + (y - 3)²

=> (x - 6)² + (y + 1)² = (x - 2)² + (y - 3)²

=> x² - 12x + 36 + y² + 2y + 1 = x² - 4x + 4 + y² - 6y + 9

=> - 12x + 36 + 2y + 1 = - 4x + 4 - 6y + 9

=> - 12x + 2y + 37 = - 4x - 6y + 13

=> (-12x + 4x) + (2y + 6y) = 13 - 37

=> -8x + 8y = -24

=> -8(x - y) = -24

=> x - y = 3

To prove, x − y = 3

Proof:-

According to the question

=> AP = BP => √(x - 6)² + (y + 1)² = √(x - 2)² + (y - 3)²

=> (x - 6)² + (y + 1)² = (x - 2)² + (y - 3)²

=> x² - 12x + 36 + y² + 2y + 1 = x² - 4x + 4 + y² - 6y + 9

=> - 12x + 36 + 2y + 1 = - 4x + 4 - 6y + 9

=> - 12x + 2y + 37 = - 4x - 6y + 13

=> (-12x + 4x) + (2y + 6y) = 13 - 37

=> -8x + 8y = -24

=> -8(x - y) = -24

=> x - y = 3

**Question 9.**

Find the coordinates of the point equidistant from three given points A(5,3), B(5,-5) and C(1,-5).

Find the coordinates of the point equidistant from three given points A(5,3), B(5,-5) and C(1,-5).

**Solution :**

**Let, the points which are equidistant from three points A(5,3), B(5,-5) and C(1,-5) be O(x,y).**

=> (OA)² = (x - 5)² + (y - 3)² = (x² + 25 - 10x) + (y² + 9 - 6y) = (x² + 25 - 10x + y² + 9 - 6y) = (x² + y² - 10x - 6y + 34)

=> (OB)² = (x - 5)² + (y + 5)² = (x² + 25 - 10x) + (y² + 25 + 10y) = (x² + 25 - 10x + y² + 25 + 10y) = (x² + y² - 10x + 10y + 50)

=> (OC)² = (x - 1)² + (y + 5)² = (x² + 1 - 2x) + (y² + 25 + 10y) = (x² + 1 - 2x + y² + 25 + 10y) = (x² + y² - 2x + 10y + 26)

According to the question

=> (OA)² = (OB)²

=> (x² + y² - 10x - 6y + 34) = (x² + y² - 10x + 10y + 50)

=> x² + y² - 10x - 6y + 34 = x² + y² - 10x + 10y + 50

=> -6y + 34 = 10y + 50

=> -16y = 16

=> y = -1

=> (OB)² = (OC)²

=> (x² + y² - 10x + 10y + 50) = (x² + y² - 2x + 10y + 26)

=> x² + y² - 10x + 10y + 50 = x² + y² - 2x + 10y + 26

=> -10x + 50 = - 2x + 26

=> -8x = -24

=> x = 3

So, the coordinates of the point is (3,1)

=> (OA)² = (x - 5)² + (y - 3)² = (x² + 25 - 10x) + (y² + 9 - 6y) = (x² + 25 - 10x + y² + 9 - 6y) = (x² + y² - 10x - 6y + 34)

=> (OB)² = (x - 5)² + (y + 5)² = (x² + 25 - 10x) + (y² + 25 + 10y) = (x² + 25 - 10x + y² + 25 + 10y) = (x² + y² - 10x + 10y + 50)

=> (OC)² = (x - 1)² + (y + 5)² = (x² + 1 - 2x) + (y² + 25 + 10y) = (x² + 1 - 2x + y² + 25 + 10y) = (x² + y² - 2x + 10y + 26)

According to the question

=> (OA)² = (OB)²

=> (x² + y² - 10x - 6y + 34) = (x² + y² - 10x + 10y + 50)

=> x² + y² - 10x - 6y + 34 = x² + y² - 10x + 10y + 50

=> -6y + 34 = 10y + 50

=> -16y = 16

=> y = -1

=> (OB)² = (OC)²

=> (x² + y² - 10x + 10y + 50) = (x² + y² - 2x + 10y + 26)

=> x² + y² - 10x + 10y + 50 = x² + y² - 2x + 10y + 26

=> -10x + 50 = - 2x + 26

=> -8x = -24

=> x = 3

So, the coordinates of the point is (3,1)

**Question 10.**

Find the ratio does the y-axis divide the line segment joining the points p(-4, 5) and Q(3, 7)?

Find the ratio does the y-axis divide the line segment joining the points p(-4, 5) and Q(3, 7)?

**Question 11.**

Find the ratio in which the line segment joining points A (a1,b1) and B (a2,b2) is divided by y-axis is ?

Find the ratio in which the line segment joining points A (a1,b1) and B (a2,b2) is divided by y-axis is ?

**Question 12.**

The length of a line segment joining A(2,-3) and B is 10 units. If the abscissa of B is 10 units, then find the ordinates of point B?

The length of a line segment joining A(2,-3) and B is 10 units. If the abscissa of B is 10 units, then find the ordinates of point B?

**Solution :**

**Given, The length of a line segment joining A(2,-3) and B is 10 units. Let, the coordinate of point B be (10,y).**

According to the question

=> √(2 - 10)² + (-3 - y)² = 10

=> √(-8)² + (9 + y² - 6y) = 10

=> 64 + 9 + y² - 6y = 100

=> y² - 6y - 27 = 0

=> y² - 9y + 3y - 27 = 0

=> y(y - 9) + 3(y - 9) = 0

=> (y + 3)(y - 9) = 0

So, the ordinates of point B are -3 and 9.

According to the question

=> √(2 - 10)² + (-3 - y)² = 10

=> √(-8)² + (9 + y² - 6y) = 10

=> 64 + 9 + y² - 6y = 100

=> y² - 6y - 27 = 0

=> y² - 9y + 3y - 27 = 0

=> y(y - 9) + 3(y - 9) = 0

=> (y + 3)(y - 9) = 0

So, the ordinates of point B are -3 and 9.

**Question 13.**

If (-1,2), (2,-1), (3,1) and (a, b) are four vertices of a parallelogram, then find the value of a and b?

If (-1,2), (2,-1), (3,1) and (a, b) are four vertices of a parallelogram, then find the value of a and b?

**Solution :**

**Let A(-1,2), B(2,-1), C(3,1) and D(a, b) are four vertices of a parallelogram.**

=> Mid-point of AC = (-1 + 3)/2, (2 + 1)/2 = (1, 3/2)

=> Mid-point of BD = (2 + a)/2, (-1 + b)/2

Since diagonals of parallelogram intersect at each other.

=> Mid-point of AC = Mid-point of BD

=> (1, 3/2) = (2 + a)/2, (-1 + b)/2

=> (2 + a)/2 = 1 and (-1 + b)/2 = 3/2

=> 2 + a = 2 and -1 + b = 3

=> a = 0 and b = 4

So, the value of A and B are 0 and 4 respetively.

=> Mid-point of AC = (-1 + 3)/2, (2 + 1)/2 = (1, 3/2)

=> Mid-point of BD = (2 + a)/2, (-1 + b)/2

Since diagonals of parallelogram intersect at each other.

=> Mid-point of AC = Mid-point of BD

=> (1, 3/2) = (2 + a)/2, (-1 + b)/2

=> (2 + a)/2 = 1 and (-1 + b)/2 = 3/2

=> 2 + a = 2 and -1 + b = 3

=> a = 0 and b = 4

So, the value of A and B are 0 and 4 respetively.

**Question 14.**

The point on the x-axis which is equidistant from (2, -5) and (-2, 9) is?

The point on the x-axis which is equidistant from (2, -5) and (-2, 9) is?

**Solution :**

**Given, points are A(2, -5) and B(-2, 9)**

Since the point is on x-axis, So its y-coordinate is 0

Let the point is C(x, 0)

According to the question

=> AC = BC

=> √(x - 2)² + (0 + 5)² = √(x + 2)² + (0 - 9)²

=> (x - 2)² + 5² = (x + 2)² + (-9)²

=> x² - 4x + 4 + 25 = x² + 4x + 4 + 81

=> -4x + 25 = 4x + 81

=> 4x + 4x = 25 - 81

=> 8x = -56

=> x = -56/8

=> x = -7

So, the point c is (-7,0)

Since the point is on x-axis, So its y-coordinate is 0

Let the point is C(x, 0)

According to the question

=> AC = BC

=> √(x - 2)² + (0 + 5)² = √(x + 2)² + (0 - 9)²

=> (x - 2)² + 5² = (x + 2)² + (-9)²

=> x² - 4x + 4 + 25 = x² + 4x + 4 + 81

=> -4x + 25 = 4x + 81

=> 4x + 4x = 25 - 81

=> 8x = -56

=> x = -56/8

=> x = -7

So, the point c is (-7,0)

**Question 15.**

The third vertex of a triangle if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.

The third vertex of a triangle if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.

**Solution :**

Let the third vertex is (x, y) and centroid of the traingle is (0,0)

According to the question

=> (x - 3 + 0)/3, (y + 1 - 2)/3 = (0, 0)

=> (x - 3)/3, (y - 1)/3 = (0, 0)

=> (x - 3)/3 = 0 and (y - 1)/3 = 0

=> x - 3 = 0 and y - 1 = 0

=> x = 3, y = 1

So, the third vertex of a triangle is (3,1)