Find the coordinates of a point that is in Quadrant IV and is √45 units away from the point (-4, 1).The coordinates are ( , )
Any point on the circle
(x+4)^2 + (y-1)^2 = 45
will do.
To make it easy, consider a 3-4-5 right triangle, scaled up to
27-36-45
So ((-4+27),(1-36)) = (31,-35) will work.
To find the coordinates of the point, we can use the distance formula between two points. The distance formula is given by:
distance = √((x2 - x1)^2 + (y2 - y1)^2),
where (x1, y1) and (x2, y2) are the coordinates of the two points.
In this case, we are given the coordinates of one point (-4, 1) and the distance (√45) to the unknown point.
Using the distance formula, we can set up the equation:
√45 = √((-4 - x)^2 + (1 - y)^2),
where (x, y) are the coordinates of the unknown point.
Simplifying the equation, we get:
45 = (-4 - x)^2 + (1 - y)^2.
Since we know that the unknown point is in Quadrant IV, we can deduce that the x-coordinate will be positive and the y-coordinate will be negative.
Expanding the equation and rearranging terms, we have:
45 = 16 + 8x + x^2 + 1 - 2y + y^2.
Rewriting the equation in the standard quadratic form:
x^2 + y^2 + 8x - 2y + (-28) = 0.
From here, we can complete the square to find the coordinates of the unknown point. However, solving this equation algebraically may not lead to simple integer solutions.
Alternatively, we can use graphing software or an online graphing tool to plot the equation:
x^2 + y^2 + 8x - 2y + (-28) = 0.
By visually observing the graph in Quadrant IV, we can identify the coordinates of the desired point where the distance from (-4, 1) is √45 units away.
Therefore, using either visual analysis or algebraic methods, we can find the approximate coordinates of the point in Quadrant IV that is √45 units away from (-4, 1).