Answer:
(18, 14)
Step-by-step explanation:
We know that C and D lie on the line AB and BC = CD = AB. Then we need to use the distance formula and equation of the line AB to find the other two coordinates.
The distance formula states that the distance between two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], the distance is denoted by: [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]. Let's find the distance between A and B:
d = [tex]\sqrt{(6-0)^2+(6-2)^2}=\sqrt{6^2+4^2} =\sqrt{36+16} =\sqrt{52} =2\sqrt{13}[/tex]
Now say the coordinates of D are (a, b). Then the distance between D and B will be twice of 2√13, which is 4√13:
4√13 = [tex]\sqrt{(6-a)^2+(6-b)^2}[/tex]
Square both sides:
208 = (6 - a)² + (6 - b)²
Let's also find the equation of the line AB. The y-intercept we know is 2, so in y = mx + b, b = 2. The slope is (6 - 2) / (6 - 0) = 4/6 = 2/3. So the equation of the line is: y = (2/3)x + 2. Since (a, b) lines on this line, we can put in a for x and b for y: b = (2/3)a + 2. Substitute this expression in for b in the previous equation:
208 = (6 - a)² + (6 - b)²
208 = (6 - a)² + (6 - (2/3a + 2))² = (6 - a)² + (-2/3a + 4)²
208 = a² - 12a + 36 + 4/9a² - 16/3a + 16 = 13/9a² - 52/3a + 52
0 = 13/9a² - 52/3a - 156
13a² - 156a - 1404 = 0
a² - 12a - 108 = 0
(a + 6)(a - 18) = 0
a = -6 or a = 18
We know a can't be negative so a = 18. Plug this back in to find b:
b = 2/3a + 2 = (2/3) * 18 + 2 = 12 + 2 = 14
So point D has coordinates (18, 14).
