NCERT MCQ Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Finding the Distance of a Point from the x-Axis
We are given a point P(2,3) and need to find its distance from the x-axis.
Step 1: Understanding Distance from the x-Axis
The x-axis is the horizontal axis in the Cartesian plane.
The distance of any point (x, y) from the x-axis is simply the absolute value of its y-coordinate.
This is because the x-axis is located at y=0, and the vertical distance from any point to this axis is determined by how far its y-value is from zero.
Step 2: Applying the Formula
For a point (x, y), the distance from the x-axis is given by:
Distance = |y|
For point P(2, 3), we substitute y = 3:
Distance = |3| = 3
Step 3: Final Answer
Thus, the distance of point P(2,3) from the x-axis is 3.

To determine the type of quadrilateral formed by the points A(9, 0), B(9, 6), C(-9, 6), and D(-9, 0), we will analyze the properties of the sides and angles using the distance formula. The distance formula between two points (x₁, y₁) and (x₂, y₂) is given by:

d = √((x₂ – x₁)² + (y₂ – y₁)²)

Step 1: Calculate the lengths of all sides
1. Length of AB:
AB = √((9 – 9)² + (6 – 0)²)
= √(0² + 6²)
= √36
= 6

2. Length of BC:
BC = √((-9 – 9)² + (6 – 6)²)
= √((-18)² + 0²)
= √324
= 18

3. Length of CD:
CD = √((-9 – (-9))² + (0 – 6)²)
= √(0² + (-6)²)
= √36
= 6

4. Length of DA:
DA = √((9 – (-9))² + (0 – 0)²)
= √((9 + 9)² + 0²)
= √324
= 18

Step 2: Analyze the side lengths
From the calculations:
– AB = CD = 6 (opposite sides are equal)
– BC = DA = 18 (opposite sides are equal)

Thus, the quadrilateral has opposite sides that are equal in length.

Step 3: Check if the angles are right angles
To confirm whether the angles are right angles, we calculate the slopes of adjacent sides and check if their product is -1 (indicating perpendicularity).

1. Slope of AB:
Slope of AB = (6 – 0) / (9 – 9) = undefined (vertical line)

2. Slope of BC:
Slope of BC = (6 – 6) / (-9 – 9) = 0 (horizontal line)

3. Slope of CD:
Slope of CD = (0 – 6) / (-9 – (-9)) = undefined (vertical line)

4. Slope of DA:
Slope of DA = (0 – 0) / (9 – (-9)) = 0 (horizontal line)

Since the slopes of adjacent sides (e.g., AB and BC, or BC and CD) indicate perpendicularity (one is vertical and the other is horizontal), all angles are right angles.

 Step 4: Conclusion
The quadrilateral has opposite sides equal and all angles are right angles. Therefore, it is a rectangle.

The distance of a point (x, y) from the origin is given by the formula:
d = √(x² + y²)

Here, the distance is given as 5 units. Therefore, we have:
√(x² + y²) = 5

Square both sides to eliminate the square root:
x² + y² = 25

This equation represents a circle with radius 5 centered at the origin.
Step 1: Analyze the third quadrant
In the third quadrant, both x and y are negative. Thus, any point (x, y) in the third quadrant must satisfy:
– x < 0 - y < 0 - x² + y² = 25 Step 2: Check if there are infinitely many solutions For any point on the circle x² + y² = 25, there are infinitely many points that satisfy this equation because it is a continuous curve. Specifically, in the third quadrant, there are infinitely many points where both x and y are negative, as long as they satisfy the circle equation. For example: - If x = -3, then y² = 25 - (-3)² = 25 - 9 = 16, so y = -4 (since y < 0 in the third quadrant). - If x = -4, then y² = 25 - (-4)² = 25 - 16 = 9, so y = -3 (since y < 0 in the third quadrant). This process can be repeated for infinitely many values of x in the range -5 ≤ x < 0, each corresponding to a unique y-value in the third quadrant. Step 3: Conclusion There are infinitely many points in the third quadrant that lie on the circle x² + y² = 25.

To determine the type of triangle formed by the points (-4, 0), (4, 0), and (0, 3), we will calculate the lengths of the sides using the distance formula. The distance formula between two points (x₁, y₁) and (x₂, y₂) is given by:
d = √((x₂ – x₁)² + (y₂ – y₁)²)

Let the points be A(-4, 0), B(4, 0), and C(0, 3).
1. Calculate the length of side AB:
AB = √((4 – (-4))² + (0 – 0)²)
= √((4 + 4)² + 0²)
= √(8²)
= √64
= 8

2. Calculate the length of side BC:
BC = √((0 – 4)² + (3 – 0)²)
= √((-4)² + 3²)
= √(16 + 9)
= √25
= 5

3. Calculate the length of side AC:
AC = √((0 – (-4))² + (3 – 0)²)
= √((0 + 4)² + 3²)
= √(4² + 3²)
= √(16 + 9)
= √25
= 5

Now that we have the side lengths:
AB = 8, BC = 5, AC = 5.

Since two sides (BC and AC) are equal, the triangle is **isosceles**. Additionally, we can check if it forms a right triangle using the Pythagorean theorem:
For a right triangle, the square of the longest side (hypotenuse) should equal the sum of the squares of the other two sides. Here, AB is the longest side.

Check: AB² = BC² + AC²
8² = 5² + 5²
64 = 25 + 25
64 ≠ 50

Since the Pythagorean theorem does not hold, the triangle is not a right triangle. It is also not equilateral because all sides are not equal. Therefore, the correct answer is isosceles triangle.

To find the value of y such that the points P(2, 4), Q(0, 3), R(3, 6), and S(5, y) form a parallelogram PQRS, we use the property that the diagonals of a parallelogram bisect each other. This means the midpoints of the diagonals PR and QS must coincide.

Step 1: Find the midpoint of diagonal PR
The formula for the midpoint of a line segment joining two points (x₁, y₁) and (x₂, y₂) is:

Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)

For diagonal PR, the endpoints are P(2, 4) and R(3, 6). The midpoint of PR is:
Midpoint of PR = ((2 + 3)/2, (4 + 6)/2)
= (5/2, 10/2)
= (5/2, 5)

Step 2: Find the midpoint of diagonal QS
For diagonal QS, the endpoints are Q(0, 3) and S(5, y). The midpoint of QS is:
Midpoint of QS = ((0 + 5)/2, (3 + y)/2)
= (5/2, (3 + y)/2)

Step 3: Equate the midpoints
Since the diagonals of a parallelogram bisect each other, the midpoints of PR and QS must be equal. Therefore: (5/2, 5) = (5/2, (3 + y)/2)

Equating the y-coordinates: 5 = (3 + y)/2

Multiply through by 2 to eliminate the denominator:
10 = 3 + y

Solve for y:
y = 10 – 3
y = 7

Step 4: Verify the solution
Substitute y = 7 into the coordinates of S(5, y), making S(5, 7). Recalculate the midpoint of QS:

Midpoint of QS = ((0 + 5)/2, (3 + 7)/2)
= (5/2, 10/2)
= (5/2, 5)

This matches the midpoint of PR, confirming that the diagonals bisect each other.
The correct answer is 7.

To solve for the value of p, we use the distance formula between two points (x₁, y₁) and (x₂, y₂):
d = √((x₂ – x₁)² + (y₂ – y₁)²)

Here, the points are (4, p) and (1, 0), and the distance is given as 5. Substituting into the formula:
5 = √((1 – 4)² + (0 – p)²)

Simplify the terms inside the square root:
5 = √((-3)² + (-p)²)
5 = √(9 + p²)

Square both sides to eliminate the square root:
5² = 9 + p²
25 = 9 + p²

Rearrange the equation to isolate p²:
p² = 25 – 9
p² = 16

Take the square root of both sides:
p = ±√16
p = ±4

Thus, the value of p can be either +4 or -4.
The correct answer is ±4.

The distance of a point (x, y) from the x-axis is given by the absolute value of its y-coordinate. This is because the x-axis is the horizontal line where y = 0, and the vertical distance between the point and the x-axis depends only on the y-coordinate.

Given the point (-6, 8), the y-coordinate is 8. The distance from the x-axis is:
Distance = |y| = |8| = 8 units

Thus, the distance of the point (-6, 8) from the x-axis is 8 units.
The correct answer is 8 units.

To find the ratio in which the x-axis divides the line segment joining the points (2, -3) and (5, 6), we use the section formula. The x-axis has the equation y = 0, so the point of division lies on the x-axis, meaning its y-coordinate is 0.

Step 1: Section formula
The section formula states that if a point (x, y) divides the line segment joining two points (x₁, y₁) and (x₂, y₂) in the ratio m:n, then:

x = (mx₂ + nx₁) / (m + n)
y = (my₂ + ny₁) / (m + n)

Here, the given points are:
(x₁, y₁) = (2, -3)
(x₂, y₂) = (5, 6)

Let the ratio be m:n. Since the point of division lies on the x-axis, its y-coordinate is 0. Using the y-coordinate formula:
y = (my₂ + ny₁) / (m + n)

Substitute y = 0, y₁ = -3, and y₂ = 6:
0 = (m(6) + n(-3)) / (m + n)

Simplify:
0 = (6m – 3n) / (m + n)

Multiply through by (m + n) (which is nonzero):
6m – 3n = 0

Rearrange to solve for the ratio m:n:
6m = 3n
m/n = 3/6
m/n = 1/2

Thus, the ratio is 1:2.
Step 2: Verify the solution
The x-axis divides the line segment in the ratio 1:2. To confirm, substitute m = 1 and n = 2 into the section formula for the y-coordinate:

y = (my₂ + ny₁) / (m + n)
y = (1(6) + 2(-3)) / (1 + 2)
y = (6 – 6) / 3
y = 0

This confirms that the point of division lies on the x-axis.
The correct answer is 1:2.

The distance of a point (x, y) from the x-axis is given by the absolute value of its y-coordinate. This is because the x-axis is the horizontal line where y = 0, and the vertical distance between the point and the x-axis depends only on the y-coordinate.

Given the point (-1, 7), the y-coordinate is 7. The distance from the x-axis is:
Distance = |y| = |7| = 7 units

Thus, the distance of the point (-1, 7) from the x-axis is 7 units.
The correct answer is 7.

To find the point on the x-axis that is equidistant from the points (-1, 0) and (5, 0), let the required point be (x, 0), since it lies on the x-axis (its y-coordinate is 0).

Step 1: Use the distance formula
The distance between two points (x₁, y₁) and (x₂, y₂) is given by:
d = √((x₂ – x₁)² + (y₂ – y₁)²)

The distances from (x, 0) to (-1, 0) and (5, 0) must be equal. Therefore:
Distance to (-1, 0) = Distance to (5, 0)
√((x – (-1))² + (0 – 0)²) = √((x – 5)² + (0 – 0)²)

Simplify both sides: √((x + 1)²) = √((x – 5)²)

Step 2: Eliminate the square roots
Square both sides to remove the square roots:
(x + 1)² = (x – 5)²

Expand both sides:
x² + 2x + 1 = x² – 10x + 25

Step 3: Simplify the equation
Cancel out x² from both sides:
2x + 1 = -10x + 25

Combine like terms:
2x + 10x = 25 – 1
12x = 24

Solve for x:
x = 24 / 12
x = 2

Step 4: Verify the solution
The point on the x-axis is (2, 0). To verify, calculate the distances from (2, 0) to (-1, 0) and (5, 0):

1. Distance to (-1, 0):
√((2 – (-1))² + (0 – 0)²) = √((2 + 1)²) = √(3²) = 3

2. Distance to (5, 0):
√((2 – 5)² + (0 – 0)²) = √((-3)²) = √(3²) = 3

Since both distances are equal, the point (2, 0) is indeed equidistant from (-1, 0) and (5, 0). The correct answer is (2, 0).