RS Aggarwal Solutions: Coordinate Geometry- 3 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

The distance between any two points P₁(x₁, y₁) and P₂(x₂, y₂) is given by the following formula:

From the question we have,
⇒ P₁(x₁, y₁) = (0, 0)…………co-ordinates of origin
⇒ P₂(x₂, y₂) = (-6, 8)…………co-ordinates of point

⇒ d = √(36 + 64 )
⇒ d = √100
⇒ d = 10 units
Therefore the distance between the point and origin is 10 units.

The distance of any point from x-axis can be determined the modulus or absolute value of the y-coordinate of that point and in similar manner the distance of any point from y-axis can be determined the modulus or absolute value of the x-coordinate of that point
The modulus of y-coordinate is taken because distance cannot be negative.
In this case the y-coordinate is 4 and hence the distance of point from x-axis is 4 units.

⇒ For the point to be equidistant, the point has to be the midpoint of the line joining the points A and B.
⇒ If P(x, y) is the midpoint of the line joining AB then By Midpoint Formula we have,

Finding x co-ordinate of midpoint:

⇒ x = 4/2
⇒ x = 2
Finding y- co-ordinate of midpoint:

⇒ y = 0
Therefore the point which is equidistant from A and B is P(2, 0).

⇒ If P(x, y) is the midpoint of the line joining AB then By Midpoint Formula we have,

Finding the value of y:


⇒ 12 = 5 + y
⇒ y = 12 – 5
⇒ y = 7
Therefore the value of y is 7.

⇒ If P(x, y) is the dividing point of the line joining AB then By Section Formula we have,

where m and n is the ratio in which the point C divides the line AB
Finding the value of k:
⇒ m = 2 and n = 3

⇒ k = 16/5
The value of k is 16/5.

The perimeter is the addition of lengths of all sides.
Let the points be A = (0, 4), B = (0, 0) and C = (3, 0).
The distance between any two points P1(x1, y1) and P2(x2, y2) is given by the following formula:

= √4²
= 4
⇒ Distance BC = 
= √3²
= 3
⇒ Distance AC = 
= √(9 + 16)
= √25
= 5
∴ Perimeter = 3 + 4 + 5
= 12
Therefore the perimeter of triangle is 12.

Since the given quadrilateral is a parallelogram, the length of parallel sides is equal.
So by distance formula,
⇒ Distance AB = 
= √(4 + 1)
= √5
⇒ Distance CD = 
= 
⇒ Distance CD = Distance AB

Squaring both sides
⇒ 5 = (x-2)² + 1
⇒ 4 = (x-2)²
Taking square root of both sides
⇒ 2 = x-2
⇒ x = 4
or
⇒ -2 = x-2
⇒ x = 0
Therefore the value of x can be 0 or 4.

Three points A, B, C are said to be collinear if,
Area of triangle formed by three points is zero
The formula of Area of Triangle of three points is given as follows:

⇒ 1/2 × {[(x + 3) × 1] – [6× -10]} = 0
⇒ x + 3 + 60 = 0
⇒ x = -63
Therefore the value of x is -63.

⇒ Formula of Area of Triangle of three points is given as follows:

= 1/2 × {[-3× -4] -0}
= 1/2 × 12
= 6
Therefore the area of a triangle in square units is 6.

⇒ Formula of Area of Triangle of three points is given as follows:

= (ab)/2
Therefore the area of the triangle is ab/2.

If P(x, y) is the midpoint of the line joining AB then By Midpoint Formula we have,

⇒ a = -8
Therefore the value of a is -8.

Distance BD is the length of one of its diagonal.
⇒ So by distance formula,


= √25
= 5
Therefore the length of diagonal is 5 units.

If P(x, y) is the dividing point of the line joining AB then By Section Formula we have,

where m and n is the ratio in which the point C divides the line AB
Finding the x-coordinate of P:

⇒ x = 9/3
⇒ x = 3
Finding the y-coordinate of P:

⇒ y = 15/3
⇒ y = 5
Therefore the coordinates of P is (3, 5).

Since the center divides the diameter into two equal halves.
⇒ Therefore by Midpoint Formula we have,

Finding the coordinates of another end of diameter:
Finding x-coordinate:

⇒ -4 = 2 + x₂
⇒ x₂ = -4-2
⇒ x₂ = -6
Finding y-coordinate:

⇒ 10 = 3 + y₂
⇒ y₂ = 10-3
⇒ y₂ = 7
Therefore the coordinates of another end of diameter are (-6, 7).

From the given diagram, we come to know
⇒ AP = PQ = QB
⇒ Therefore the point P divides the line internally in the ratio 1:2 and Q divides the line in the ratio 2: 1
⇒ Then by section formula the y-coordinate of point Q which divide the line AB is given as

⇒ y = -12/3
⇒ y = -4
Therefore the value of y is -4.

⇒ Therefore by Midpoint Formula we have,

Finding the coordinates of the end of A:
⇒ Finding x-coordinate:

⇒ x₂ = 2
Finding y-coordinate:

⇒ 8 = 3 + y₂
⇒ y₂ = 8-3
⇒ y₂ = 5
Therefore the coordinates of the end of A are (2, 5).

If P(x, y) is the dividing point of the line joining AB then By Section Formula we have,

⇒ where m and n is the ratio in which the point C divides the line AB
Finding the x-coordinate of P:

⇒ x = 16/3
Finding the y-coordinate of P:

⇒ y = -11/3
Therefore the coordinates of P is (16/3, -11/3).
Since in fourth quadrant x-coordinate is positive and y-coordinate is negative.
Therefore the point P lies in the fourth quadrant.

= √(25 + 144 )
= √(169 )
= 13
⇒ Distance 2AB = 2×13
= 26 units.
Therefore the distance 2AB is 26 units.

⇒ Point on x-axis means its y-coordinate is zero.
⇒ Let the point be P(x, 0)
Using the distance formula,

⇒ Distance AP = Distance BP
⇒ (x-7)² + (0-6)² = (x + 3)² + (0-4)²
⇒ x² + 49-14x + 36 = x² + 9 + 6x + 16
⇒ 49-9 + 36-16 = 6x + 14x
⇒ 40 + 20 = 20x
⇒ x = 60/20
x = 3
Therefore the coordinate of P is (3,0).

The distance of any point from x-axis can be determined the modulus or absolute value of the y-coordinate of that point and in a similar manner, the distance of any point from y-axis can be determined the modulus or absolute value of the x-coordinate of that point
The modulus of y-coordinate is taken because distance cannot be negative.
In this case, the y-coordinate is 4 and hence the distance of the point from x-axis is 4 units.

⇒ Let the ratio be k:1.
⇒ Then by section formula the coordinates of point which divide the line AB is given as

⇒ Since the point lies on x-axis its y-coordinate is zero.

⇒ 6k = 3
⇒ k = 1/2
Therefore the ratio in which x-axis divide the line AB is 1:2.

⇒ Let the ratio be k:1.
⇒ Then by section formula the coordinates of point which divide the line AB is given as

⇒ Since the point lies on y-axis its x-coordinate is zero.

⇒ 8k = 4
⇒ k = 1/2
Therefore the ratio in which x-axis divide the line AB is 1:2.

∴ by Midpoint Formula we have,

Finding value of b:

⇒ 2 = 2b + 4
⇒ 2 – 4 = 2b
⇒ b = -2/2
⇒ b = -1
Therefore the value of b is -1.

⇒ Let 2x + y = 4 ………….. (1)
Finding the equation of line formed by AB:
Finding slope:


⇒ m = 9
The equation of line AB:
⇒ y – y1 = m×(x-x1)
⇒ y-(-2) = 9×(x-2)
⇒ y + 2 = 9x – 18
⇒ 9x – y = 20……(2)
When we solve the two equations simultaneously, we get point of intersection of two lines.
⇒ Adding (1) and (2)
⇒ 11x = 24
⇒ x = 24/11
⇒ Substituting the value of x in (1)
⇒ 2×24/11 + y = 4
⇒ y = 4 – 48/11
⇒ y = -4/11
let us assume the line divides the segment AB in the ratio k:1
Then by section formula, the coordinates of point which divide the line AB is given as

Since we know x-coordinate of the point

⇒ 33k + 22 = 24k + 24
⇒ 9k = 2
⇒ k = 2:9
Therefore the line 2x + y -4 = 0 divides the line segment AB into the ratio 2:9.

Since the AD is median, it divides the line BC into two equal halves. So D acts as the midpoint of line BC.
If D(x, y) is the midpoint of the line joining BC then By Midpoint Formula we have,

Finding x co-ordinate of midpoint:

⇒ x = 7/2
⇒ x = 7/2
Finding y- co-ordinate of midpoint:

⇒ y = 9/2
Therefore the point which is equidistant from A and B is P(7/2,9/2).

Let P(x, y) be the centroid of the triangle
⇒ Finding the x-coordinate of P:

⇒ x = 12/3
⇒ x = 4
Finding the y-coordinate of P:

⇒ y = 0
Therefore the coordinates of P are (4, 0).

Finding the x-coordinate of C:

⇒ x = -4
Finding the y-coordinate of P:

⇒ -9 = 6 + y
⇒ y = -15
Therefore the coordinates of P are (-4, -15).

The distance between any two points P₁(x₁, y₁) and P₂(x₂, y₂) is given by the following formula:


=√8²
= 8
⇒ Distance BC = 

= 5
⇒ Distance AC = 
= √(9 + 16)
= √25
= 5
Since the length of two sides is equal, given triangle is an isosceles triangle.

The distance between any two points P1(x1, y1) and P2(x2, y2) is given by the following formula:


= √(25 + 9)
= √34
⇒ Distance BC = 
= √(64 + 4)
= √68
⇒ Distance AC = 
= √(9 + 25)
= √34
Since the length of two sides is equal, given triangle is an isosceles triangle.
⇒ The given triangle also satisfy Pythagoras Theorem in following way:
BC² = AC² + AB²
Therefore the given triangle is also right-angled triangle.

Three points A, B, C are said to be collinear if,
Area of triangle formed by three points is zero
The formula of Area of Triangle of three points is given as follows:
Area, 

⇒ 1/2 × {[-3k + 21] – [-3 + k]} = 0
⇒ -4k + 21 + 3 = 0
⇒ 4k = 24
⇒ k = 6
Therefore the value of k is 6.

Three points A, B, C are said to be collinear if,
Area of triangle formed by three points is zero
⇒ Formula of Area of Triangle of three points is given as follows:


⇒ 1/2 × {[-b×1] – [-a×2]} = 0
⇒ 2a-b = 0
⇒ 2a = b
Hence Proved.

The formula of Area of Triangle of three points is given as follows:
Area, 

= 1/2 × {[-4× -4] - 0}
= 8 sq. units
Therefore the area of the triangle is 8 sq. units.

Distance BD is the length of one of its diagonal.
So by distance formula,
Distance AB = 
= √(25 + 9)
= √34 units
Therefore the length of diagonal is √34 units.

The distance between any two points P1(x1, y1) and P2(x2, y2) is given by the following formula:

⇒ From the question we have,
⇒ A = (4, p)
⇒ B = (1, 0)
⇒ d = 5

⇒ Squaring both sides
⇒ 25 = (-3)² + p²
⇒ 25 = 9 + p²
⇒ p² = 25 – 9
⇒ p² = 16
⇒ p = ±4
Therefore the value of p is ±4.