1)Find the length of the perpendicular drawn on x axis from (5, - 7) 2)What is the co-ordinate of the point on the negative side of x-axis which is at a distance of 3 units from the origin?
1)Find the length of the perpendicular drawn on x axis from (5, - 7)
2)What is the co-ordinate of the point on the negative side of x-axis which is at a distance of 3 units from the origin?
To find the length of the perpendicular drawn on x-axis from (5, -7), we need to first visualize the point and the perpendicular.
The point (5, -7) is located in the fourth quadrant, which means that the perpendicular drawn from it on the x-axis will make a right angle with the x-axis and will be directed downwards.
To find the length of the perpendicular, we need to find the y-coordinate of the point where the perpendicular intersects the x-axis. Since the perpendicular is directed downwards, we can subtract the absolute value of the y-coordinate of the given point from 0 to get the length of the perpendicular.
The y-coordinate of the given point is -7, so the length of the perpendicular is |-7 - 0| = 7 units.
Therefore, the length of the perpendicular drawn on x-axis from (5, -7) is 7 units.
To find the coordinate of the point on the negative side of x-axis which is at a distance of 3 units from the origin, we can use the distance formula:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
where (x₁, y₁) = (0, 0) is the origin, and (x₂, y₂) is the point we are looking for.
Since we know that the point is on the negative side of the x-axis, we can write its coordinates as (x, -y), where x is a negative number and y is a positive number.
Substituting these coordinates into the distance formula and simplifying, we get:
3 = √[(x - 0)² + (-y - 0)²]
9 = x² + y²
Since we are looking for a point with a distance of 3 units from the origin, we can substitute 3 for d and solve for y in terms of x:
9 = x² + y²
9 = x² + (-y)²
9 = x² + y²
y² = 9 - x²
y = √(9 - x²)
Now we can substitute this expression for y into the coordinate expression for the point:
(x, -y) = (x, -√(9 - x²))
Since we know that x is negative, we can simplify this expression as:
(x, -y) = (-√(9 - x²), √(9 - x²))
Therefore, the coordinate of the point on the negative side of x-axis which is at a distance of 3 units from the origin is (-√6, √3).
Answers & Comments
Shivansh Saini
1)Find the length of the perpendicular drawn on x axis from (5, - 7)
2)What is the co-ordinate of the point on the negative side of x-axis which is at a distance of 3 units from the origin?
To find the length of the perpendicular drawn on x-axis from (5, -7), we need to first visualize the point and the perpendicular.
The point (5, -7) is located in the fourth quadrant, which means that the perpendicular drawn from it on the x-axis will make a right angle with the x-axis and will be directed downwards.
To find the length of the perpendicular, we need to find the y-coordinate of the point where the perpendicular intersects the x-axis. Since the perpendicular is directed downwards, we can subtract the absolute value of the y-coordinate of the given point from 0 to get the length of the perpendicular.
The y-coordinate of the given point is -7, so the length of the perpendicular is |-7 - 0| = 7 units.
Therefore, the length of the perpendicular drawn on x-axis from (5, -7) is 7 units.
To find the coordinate of the point on the negative side of x-axis which is at a distance of 3 units from the origin, we can use the distance formula:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
where (x₁, y₁) = (0, 0) is the origin, and (x₂, y₂) is the point we are looking for.
Since we know that the point is on the negative side of the x-axis, we can write its coordinates as (x, -y), where x is a negative number and y is a positive number.
Substituting these coordinates into the distance formula and simplifying, we get:
3 = √[(x - 0)² + (-y - 0)²]
9 = x² + y²
Since we are looking for a point with a distance of 3 units from the origin, we can substitute 3 for d and solve for y in terms of x:
9 = x² + y²
9 = x² + (-y)²
9 = x² + y²
y² = 9 - x²
y = √(9 - x²)
Now we can substitute this expression for y into the coordinate expression for the point:
(x, -y) = (x, -√(9 - x²))
Since we know that x is negative, we can simplify this expression as:
(x, -y) = (-√(9 - x²), √(9 - x²))
Therefore, the coordinate of the point on the negative side of x-axis which is at a distance of 3 units from the origin is (-√6, √3).
Verified answer
EXPLANATION.
Question = 1.
Find the length of the perpendicular drawn on x - axis from (5, -7).
As we know that,
Distance formula.
⇒ d = √[(x₁ - x₂)² + (y₁ - y₂)²].
Using this formula in this question, we get.
Coordinates on x - axis = (5, 0).
⇒ d = √[(5 - 5)² + (- 7 - 0)²].
⇒ d = √[(0)² + (-7)²].
⇒ d = √(0 + 49).
⇒ d = √49.
⇒ d = 7.
∴ The length of the perpendicular drawn on x - axis from (5, -7) is 7.
Question = 2.
What is the Co-ordinate of the point on the negative side of x - axis which is at a distance of 3 units from the origin.
As we know that,
Coordinates axis of x - axis on negative side is = (- x, 0).
Where x is the distance from the origin.
∴ The Coordinate is = (- 3, 0).