I need help with two math problems.
1. A triangle has vertices (1, 4), (1, 1), and (-3, 1). The triangle is dilated by a scale factor of 2, then translated 5 units up, and then rotated 90 degrees counterclockwise about the origin. What are the vertices of the image of the triangle?
2. Use matrices to solve the system.
2x+3y=4
5x+8y=11
I appreciate the help!
1.
The transformations are:
e(O,2) : dilation about origin of 2 units
t(0,5) : translation 5 units up
r(y,-x) : rotation counter-clockwise 90°
So
(x,y)-> (2x,2y) -> (2x, 2y+5) -> (2y+5, -2x)
So combining it all:
(x,y) -> (2y+5, -2x)
2.
Using Cramer's rule:
Δ=8*2-3*5=1
x=(4*8-11*3)/Δ=-1
y=(2*11-5*4)/Δ=2
Check:
2(-1)+3*2=4
5(-1)+8*2=11
ok.
I'm still confused on the first one. Thank you. I appreciate it.
I need help
Of course! Let's go through each problem step by step and explain how to find the solutions.
1. To find the vertices of the image of the triangle after the dilation, translation, and rotation, you can follow these steps:
a) Dilation: Multiply the coordinates of each vertex by the scale factor to get the new coordinates. In this case, the scale factor is 2, so the new coordinates of each vertex would be (2 * x-coordinate, 2 * y-coordinate).
Vertex 1: (1, 4) -> (2 * 1, 2 * 4) = (2, 8)
Vertex 2: (1, 1) -> (2 * 1, 2 * 1) = (2, 2)
Vertex 3: (-3, 1) -> (2 * -3, 2 * 1) = (-6, 2)
b) Translation: Add the specified values for translation to each coordinate of the vertices.
Vertex 1: (2, 8) -> (2, 8 + 5) = (2, 13)
Vertex 2: (2, 2) -> (2, 2 + 5) = (2, 7)
Vertex 3: (-6, 2) -> (-6, 2 + 5) = (-6, 7)
c) Rotation: To rotate the vertices counterclockwise by 90 degrees, you can swap the x and y coordinates and negate the new x-coordinate.
Vertex 1: (2, 13) -> (13, -2)
Vertex 2: (2, 7) -> (7, -2)
Vertex 3: (-6, 7) -> (7, 6)
Therefore, the vertices of the image of the triangle are (13, -2), (7, -2), and (7, 6).
2. To solve the system of equations using matrices, we can represent the equations in matrix form and then solve for the variables. Here's how you can do it:
a) Set up the coefficient matrix (A) and the constant term matrix (B) for the system of equations. Each matrix has the coefficients or constants arranged in rows.
A = [2, 3; 5, 8] (2x2 matrix)
B = [4; 11] (2x1 matrix)
b) To find the solution matrix (X) containing the values of x and y, you can apply the matrix equation: AX = B, where X is the unknown matrix.
AX = B
[2, 3; 5, 8] * [x; y] = [4; 11]
c) To solve for X, we can multiply both sides of the equation by the inverse of A.
X = A^(-1) * B,
where A^(-1) is the inverse of matrix A.
d) Calculate the inverse of matrix A. If A is invertible, its inverse denoted as A^(-1) exists.
A^(-1) = 1 / (ad - bc) * [d, -b; -c, a]
In our case,
ad - bc = (2 * 8) - (3 * 5) = 16 - 15 = 1,
so the inverse of A is A^(-1) = [8, -3; -5, 2].
e) Multiply the inverse of A with matrix B to get the solution matrix X.
X = A^(-1) * B
X = [8, -3; -5, 2] * [4; 11]
Multiply the elements of the matrices to find the values of x and y.
X = [8 * 4 + (-3) * 11; -5 * 4 + 2 * 11]
X = [32 - 33; -20 + 22]
X = [-1; 2]
Therefore, the solution to the system of equations is x = -1 and y = 2.
I hope this explanation helps you solve the math problems! Let me know if you have any further questions.