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The point on the x-axis which is equidistant from points (-1, 0) and (5, 0) is
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 -Read more
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:
b) (2, 0)
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.
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Point (x, y) is at distance of 5 units from the origin. How many such points lie in the third quadrant?
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. StepRead more
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.
The correct answer is:
d) infinitely many
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.
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Find the ratio in which the line segment joining (2, -3) and (5, 6) is divided by x-axis
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 ifRead more
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:
a) 1:2
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.
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If P(2, 4), Q(0, 3), R(3, 6) and S(5, y) are the vertices of a parallelogram PQRS, then the value of y is
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 TheRead more
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:
a) 7
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.
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The distance of the point (-1, 7) from x axis is
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 isRead more
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:
b) 7
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-7/