The green line is the set of all points having a y-coordinate of 6.
The green circle, with center (1,2) and radius 17, is the set of all points
whose distance from point (1,2) is 17. So the 2 points they have in common
are the only two points having both properties. So the answer will be those
two points.
B is 6 units above the x-axis and A is 2 units above the x-axis, so the
distance AB = 6-2 = 4.
We have AB=4 and AC=17, .
B is 1 unit farther to the right of the y-axis, so is C, and therefore
That makes C' one unit less to the left of the y-axis, so
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Distance formula:
Let the point (x,y) [where y=6, which means the point (x,6)] have distance
17 from point (1,2).
Then
Square both sides:
Thus the two points are
and
Edwin
.
Find all points having a y-coordinate of -6 whose distance from the point (1,2) is 17.
(a)By using the Pythagorean Theorem.
(b)By using the distance formula.
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In the post by Edwin, the solution and the answer are incorrect.
In his solution, Edwin mistakenly used y-coordinate 6 instead of the given -6.
His plot is inadequate, too.
I came to bring a correct solution.
Let's call the given point A = (1,2).
All points having y-coordinate of -6, lie on the horizontal line y = -6.
Vertical distance from the point (1,2) to this line (or simply the distance) is 6+2 = 8 units.
This distance is the length of the perpendicular from this point (1,2) to the line y= -6.
We want to find points C on the line y= -6 such that the distance from A to C is 17 units.
Draw the perpendicular AB from A to line y= -6. The length of this perpendicular is 8 units.
The coordinates of B are (1,-6).
The triangle ABC is a right-angled triangle.
Its diagonal AC has the length of 17 units; its leg AB is of 8 units.
Hence, the leg BC along the line y = -6 is (Pythagoras)
= = = 15 units.
Thus possible points C are (1+15,-6) = (16,-6) or (1-15,-6) = (-14,-6).
ANSWER. There are two such points C = (16,-6) and C' = (-14,-6).
Solved (correctly).