Coordinate Geometry I Class 10 Math's I Chapter 7

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?

Let us take a point A has Cartesian coordinate (2, 5). Which we want to locate 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 = √[(x₂ – x₁)² + (y₂ – y₁)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)

n C lying between A and B 

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 Formula
Three 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)

Area of triangle

Three points A, B and C are said to be collinear if 
Area of triangle ABC = 0
(1/2) [x${}_1$$1$(y${}_2$$2$ − y${}_3$$3$) + x${}_2$$2$(y${}_3$$3$ − y${}_1$$1$) + x${}_3$$3$(y${}_1$$1$ − y${}_2$$2$)] = 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.

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.

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.

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.

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.

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.

Question 4.
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.

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

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.

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).

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).

Question 7.
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]

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.

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

Question 9.
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)

Question 10.
Find the ratio does the y-axis divide the line segment joining the points p(-4, 5) and Q(3, 7)?

Solution :

Given points are p(-4, 5) and Q(3, 7).
Let Y axis devides line joning p(-4, 5) and Q(3, 7)in ratio k:1.
Therefore, X- cordinate will be zero.
Using section formula:

=>x=(-4x1)+(3k)/(k+1)
=>0=(-4x1)+(3k)/(k+1)
=>0=(-4x1)+(3k)
=>3k=4
=>k=4/3
Ration will be 4:3.

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 ?

Solution :

Given points are A (a1,b1) and B (a2,b2).
Let Y axis devides line joning points A (a1,b1) and B (a2,b2) in ratio k:1.
Therefore, X- cordinate will be zero.
Using section formula:

=>x=(a1x1)+(a2k)/(k+1)
=>0=(a1)+(a2k)/(k+1)
=>0=(a1)+(a2k)
=>-a1/a2=k
Ration will be -a1:a2.

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?

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.

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?

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.

Question 14.
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)

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.

Solution :

Given, two of its vertices are at (-3,1) and (0,-2)
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)