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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).
The weight of the new student is 27 kg.
Average weight
= total weight ÷total number of students
<h3>1) Define variables</h3>Let the total weight of the 35 students be y kg and the weight of the new student be x kg.
<h3>2) Find the total weight of the 35 students</h3><u></u>
y= 35(45)
y= 1575 kg
<h3>3) Write an expression for average weight of students after the addition of the new student</h3>New total number of students
= 35 +1
= 36
Total weight
= total weight of 35 students +weight of new students
= y +x
36(44.5)= 1575 +x
1602= x +1575
<em>Subtract 1575 from both sides:</em>
x= 1602 -1575
x= 27
Thus, the weight of the new student is 27 kg.