find the coordinates of the points which are simultaneously 6 units from the points (6,6) and equidistant from the x and y axis. There are two ways to do this. I would like them both.
method 1:
to be equidistant from both the x and the y axes, point would have to lie on y = x
So the point would have to have the form (x,x)
distance from (6,6) has to be 6
√( (x-6)^2 + (x-6)^2 = 6
√( 2x^2 - 24x + 72) = 6
square both sides
2x^2 - 24x + 72 = 36
x^2 - 12x + 18 = 8
complete the square:
x^2 - 12x + ... = -18 + ...
x^2 - 12x + 36 = -18+36
(x-6)^2 = 18
x-6 = ± 18
x = 6 ± √18
there are two such points,
(6+√18 , 6+√18) and (6-√18 , 6-√18)
method 2
To be equidistant from (6,6) and have that length be 6, would require our points to lie on a circle with radius 6 and centre (6,6)
equation:
(x-6)^2 + (y-6)^2 = 36
but it must also lie on y = x
so solving y = x with the circle --->
(x-6)^2 + (x-6)^2 = 36
2x^2 - 24x + 72 = 36
we reach the same equation as in method 1
To solve this problem, we can approach it using two different methods: geometry and algebra. Let's go through both approaches:
Method 1: Geometry
1. Start by drawing a coordinate system with the x and y axes intersecting at the origin (0,0).
2. Plot the point (6,6) on the coordinate plane.
3. Draw two circles centered at (6,6) with a radius of 6 units. These circles represent the points that are 6 units away from (6,6).
4. Now, we want to find the points that are equidistant from the x and y axes. It means these points lie on the angle bisector of the 45-degree angle formed by the x and y axes.
5. Draw the angle bisector passing through the origin (0,0) dividing the 45-degree angle in half.
6. The intersection points of the angle bisector and the circles are the points we are looking for. Label them as A and B.
Method 2: Algebra
1. Let's consider a point (x,y) that is equidistant from the x and y axes.
2. For this point, the distance to the x-axis is equal to the distance to the y-axis. Hence, |y - 0| = |x - 0|.
3. Simplifying the equation, we get |y| = |x|.
4. Since coordinates can be positive or negative, we have the following two cases:
a) If y = x, then the point lies on the line y = x.
b) If y = -x, then the point lies on the line y = -x.
5. Now, let's find the points that are 6 units away from (6,6).
6. Substitute the values of x and y into the distance formula: √((x - 6)^2 + (y - 6)^2) = 6.
7. Solve the equation to find the values of x and y.
By applying either method, we can find the coordinates of the points that satisfy both conditions.