PROBLEM #1: Find the lengths of PR and QS.
1 answer:
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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 and
, the distance is denoted by:
. Let's find the distance between A and B:
d =
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 =
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).