Find all the points having an x coordinate of 2 whose distance from the point (-1,-5) is 5.
so far i have
5=sqrt((-1-2)^2+(-2-y)^2)
5=(-3)^2
how do i simplify the rest of the equation so i can find x. do i foil it out? thanks.
Geometrically, you are looking for the intersection of a vertical line (x=2) with a circle centred at (-1,-5) with a radius of 5.
Line L1: x=2
Circle C1: (x-(-1))²+(y-(-5))²=5²
(x+1)²+(y+5)²=25
Substitute x=2 in C1:
(2+1)²+(y+5)²=25
(y+5)²=25-3²=16=4²
(y+5) = ±4
So there are two solutions, y=-9 or y=-1
thank you!!
You're welcome.
Do give a check on the distance between the centre and the intersection points to complete the question.
To simplify the equation and solve for x, you need to square both sides of the equation:
5^2 = (-3)^2
This gives:
25 = 9
Since 25 ≠ 9, the equation is not true. Therefore, there are no points with an x-coordinate of 2 that have a distance of 5 from the point (-1, -5).
If you want to find the points that are 5 units away from (-1, -5), you can use the distance formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Plugging in the values:
5 = √((-1 - 2)^2 + (-5 - y)^2)
5 = √((-3)^2 + (-5 - y)^2)
Now, square both sides of the equation to eliminate the square root:
25 = (-3)^2 + (-5 - y)^2
25 = 9 + (-5 - y)^2
Subtract 9 from both sides of the equation:
16 = (-5 - y)^2
Take the square root of both sides to remove the square:
±4 = -5 - y
Now, solve for y:
y = -5 - 4
y = -9
So, the points that have an x-coordinate of 2 and are 5 units away from (-1, -5) are (2, -9) and (2, 1).