Answer:
(3, 3) and (15, 15)
Step-by-step explanation:
The points equidistant from the given point and the y-axis lie on the parabola that has (3,6) as its focus and the y-axis as its directrix. The equation for that can be simplified from ...
(x -3)^2 +(y -6)^2 = x^2
-6x +9 +y^2 -12y +36 = 0 . . . . . subtract x^2, eliminate parentheses
We can find the points that lie on the line y=x (equidistant from both axes) by substituting y for x or vice versa. Then we have the quadratic ...
x^2 -18x +45 = 0 . . . . substitute x for y and collect terms
(x -3)(x -15) = 0 . . . . factor it
x = 3 or 15
So, the points of interest are (x, y) = (3, 3) and (x, y) = (15, 15).
Answer:
9 units
Step-by-step explanation:
By the midsegment theorem,
Since the ratio then
and
Draw two heights LE and MF. Quadrilateral ELMF is a rectangle, then EF = LM = 6 units.
Trapezoid KLMN is isosceles trapezoid (KL = MN), then KE = FN and
Consider right triangle KLE. By the Pythagorean theorem,
