Points, A(3, −1), B(2, 1) and C(−4, 0) are corners of an isosceles trapezium ABCD. Find coordinates of D if AB is parallel to DC.
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To find the coordinates of point D, we need to understand the concept of an isosceles trapezium and the properties of parallel lines.
An isosceles trapezium is a quadrilateral with two opposite sides that are parallel. In this case, we are given that AB is parallel to DC.
Let's first calculate the slope of AB. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
m = (y2 - y1) / (x2 - x1)
Using the coordinates of points A and B, we can calculate the slope of AB as follows:
m_AB = (1 - (-1)) / (2 - 3) = 2 / (-1) = -2
Since AB is parallel to DC, the slope of DC will also be -2.
Now, let's find the equation of the line DC. We can use the point-slope form of a linear equation, which is given by:
y - y1 = m(x - x1)
We will use point C (-4, 0) and the slope -2 to find the equation of line DC.
y - 0 = -2(x - (-4))
y = -2(x + 4)
Now, we want to find the coordinates of point D, which lies on the line DC. Let's substitute the coordinates of D as (x, y) into the equation of line DC:
y = -2(x + 4)
Since D is also one of the points of the isosceles trapezium, it means that the distance between D and C is equal to the distance between A and B.
Using the distance formula, the distance between A(3, -1) and B(2, 1) is given by:
d_AB = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((2 - 3)^2 + (1 - (-1))^2)
= sqrt((-1)^2 + (2)^2)
= sqrt(1 + 4)
= sqrt(5)
The distance between D(x, y) and C(-4, 0) is also sqrt(5) since AD is parallel to BC and AB is parallel to DC.
Using the distance formula again, we can write the equation for the distance between D and C as:
d_DC = sqrt((x - (-4))^2 + (y - 0)^2)
= sqrt((x + 4)^2 + y^2)
Since AB is parallel to DC, the distances d_AB and d_DC are equal. So, we can set the expressions equal to each other and solve for x:
sqrt(5) = sqrt((x + 4)^2 + y^2)
Squaring both sides of the equation:
5 = (x + 4)^2 + y^2
Expanding and simplifying the equation:
5 = x^2 + 8x + 16 + y^2
Rearranging terms and simplifying:
x^2 + 8x + (y^2 - 11) = 0
This is the equation of a quadratic. We need to find the values of x and y that satisfy this equation, which will give us the coordinates of point D.
Solving a quadratic equation often involves factoring, completing the square, or using the quadratic formula. However, in this case, we can see that the given coordinates A, B, C, and D form a parallelogram.
Since AB is parallel to DC, and the opposite sides of a parallelogram are equal, we can determine that AD is equal in length to BC.
Therefore, if A(3, -1) and B(2, 1) have a distance of sqrt(5), then the coordinates of D are obtained by moving sqrt(5) units in the opposite direction of AB (which is the vector (-1, 2) that connects points A and B).
So the coordinates of D will be:
x = 3 - 1 = 2
y = -1 - 2 = -3
Therefore, the coordinates of point D are (2, -3).
AB is y = -2x+5
CD is y = -2x-8
the perpendicular bisector of AB meets CD at P=(-2.7,-2.6)
since C = P+(-1.3,2.6) that means
D = P+(1.3,-2.6) = (-1.4,-5.2)