Find all points on the x-axis that are 10 units from the point (4,-5).
X^2=72
Well, if we want to find all points on the x-axis that are 10 units from the point (4, -5), we can start by imagining a circle with a radius of 10 units centered at (4, -5). Now, since we're only interested in the x-axis, we can ignore the y-value of the points and focus solely on the x-value.
Now, let's think about it for a moment... If a point lies on the x-axis, that means its y-value is zero. So, basically, we just need to find all the x-values that are 10 units away from 4.
Hmm... in order to find those x-values, weneed to think about the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2). In our case, y2 - y1 is zero since all the points lie on the x-axis. So, our distance formula simplifies to d = sqrt((x2 - x1)^2).
Since we know that the distance is 10 and (x2, y2) is any point on the circle (let's call it (x, 0)), we can rewrite the formula as 10 = sqrt((x - 4)^2).
Alright, now it's time for a little math magic. We'll square both sides of the equation to get rid of the square root and simplify it. 10^2 = ((x - 4)^2.
Squaring both sides gives us 100 = (x - 4)^2.
Now, let's take the square root of both sides: sqrt(100) = sqrt((x - 4)^2).
Simplifying it gives us 10 = (x - 4).
Phew! Finally, the equation is in a form where x is alone. So, let's solve for x:
x - 4 = 10.
x = 10 + 4 = 14.
So, the only point on the x-axis that is 10 units away from the point (4, -5) is (14, 0). It's the only x-value that satisfies our distance equation.
To find all the points on the x-axis that are 10 units away from the point (4, -5), we need to determine the x-coordinate of these points.
Given that the y-coordinate of the points on the x-axis is always 0, we can draw a right triangle with legs of length 10 and 5, and the hypotenuse connecting the point (4, -5) to the desired point on the x-axis.
To find the x-coordinate, we can use the Pythagorean theorem:
c^2 = a^2 + b^2
Here, a = 5 (the vertical distance from -5 to 0) and c = 10 (the distance from (4, -5) to the desired point on the x-axis).
Let's substitute the values into the equation:
10^2 = a^2 + 5^2
100 = a^2 + 25
a^2 = 100 - 25
a^2 = 75
To solve for a, we take the square root of both sides:
a = √75
Simplifying the square root:
a ≈ √(25*3)
a ≈ 5√3
Since we are looking for the x-coordinates of the points, the point (4 - 5√3, 0) and (4 + 5√3, 0) are the points on the x-axis that are 10 units away from (4, -5).
To find all the points on the x-axis that are 10 units from the point (4, -5), we can follow the following steps:
Step 1: Recall that any point on the x-axis has a y-coordinate of 0.
Step 2: Use the distance formula to calculate the distance between the point (4, -5) and any point on the x-axis. The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, since the y-coordinate of the x-axis points is 0, we have:
d = sqrt((x - 4)^2 + (-5 - 0)^2)
Step 3: Set up an equation using the distance formula and solve for x:
10 = sqrt((x - 4)^2 + (-5)^2)
Squaring both sides of the equation, we get:
100 = (x - 4)^2 + 25
Simplifying further, we have:
75 = (x - 4)^2
Step 4: Take the square root of both sides to solve for x:
sqrt(75) = x - 4
Now, we have two possible solutions for x:
x - 4 = sqrt(75) or x - 4 = -sqrt(75)
Solving each equation separately gives us:
x = sqrt(75) + 4 or x = -sqrt(75) + 4
Step 5: Simplify the solutions:
x = sqrt(75) + 4 or x = 4 - sqrt(75)
Therefore, the points on the x-axis that are 10 units from the point (4, -5) are (sqrt(75) + 4, 0) and (4 - sqrt(75), 0).
let the point be (x,0)
distance = √(x-4)^2 + (0+5)^2 ) = 10
square both sides:
(x-4)^2 + 25 = 100
x^2 - 8x = 75
completing the square
x^2 - 8x + 16 = 75+16
(x-4)^2 = 91
x - 4 = ±√91
x = 4 ± √91
or
consider a circle with centre (4,-5) and radius 10, then find the x-intercepts
(x-4)^2 + (y+5)^2 = 100
for x-intercepts, let y = 0
(x-4)^2 + 25 = 100 , same equation as above