Coordinate representation of linear and affine transformations of the plane. In the problems, the coordinate system is assumed to be a general Cartesian one. (12.37-12.52)

a homothety with center at the origin and ratio $k$

a homothety with center at the point $M\left(x_{0}, y_{0}\right)$ and ratio $k$

a point reflection with respect to the point $M\left(x_{0}, y_{0}\right)$

a translation by the vector $\mathbf{a}(\alpha , \beta )$.

of the points

(1) $O(0, 0)$;

(2) $E_{1}(1, 0)$;

(3) $E_{2}(0, 1)$;

(4) $E(1, 1)$;

(5) $M(- 1, 5)$;

of the lines

(1) $y = 0$;

(2) $x = 0$;

(3) $x - y + 1 = 0$;

(4) $x - y - 1 = 0$;

(5) $2x + 3y = 7$.

of the points

(1) $O(0, 0)$;

(2) $A(- 1, 2)$;

(3) $B(4, - 5)$

of the lines

(1) $y = 0$;

(2) $x = 0$;

(3) $x + y - 1 = 0$;

(4) $x - y - 1 = 0$;

(5) $x - y + 1 = 0$.

$A(1, 0), B(0, 1), C(1, 1), A^{ * }(- 3, 5), B^{ * }(4, - 3), C^{ * }(0, 0)$

$A(3 / 7, 1), \quad B(1, 1 / 4), \quad C(2, - 1), \quad A^{ * }(- 4, 2), \quad B^{ * }(- 1, 6)$, $C^{ * }(4, 13)$

$A(1, 0), \quad B(- 1 / 2, \sqrt{3} / 2), \quad C(- 1 / 2, - \sqrt{3} / 2), \quad A^{ * } = B$, $B^{ * } = C, C^{ * } = A$

$A(1, 2), B(- 7, 4), C(3, - 6); A^{ * }, B^{ * }, C^{ * }$ are the midpoints of the sides of the triangle $ABC$ opposite the vertices $A, B, C$ respectively.

$A(1, 4), B(- 2, 1), C(0, 3), A^{ * }(0, 0), B^{ * }(1, 0), C^{ * }(0, 4)$

$A(- 2, 0), B(2, - 1), C(0, 4), A^{ * }(- 2, 1), B^{ * }(2, 1), C^{ * }(0, 1)$

$A(2, 0), \quad B(3, - 1), \quad C(4, - 2), \quad A^{ * }(2, 1), \quad B^{ * }(- 2, - 1)$, $C^{ * }(- 6, - 3)$

$A(0, 0), \quad B(- 1, 2), \quad C(1, - 2), \quad A^{ * }(- 1, - 1), \quad B^{ * }(0, 0)$, $C^{ * }(1, 1)$

$x^{ * } = 7x - 3y, \quad y^{ * } = x + y$

$x^{ * } = - 5x + y, \quad y^{ * } = 6x$

$x^{ * } = - 5x + y, \quad y^{ * } = 6x + 1$

$x^{ * } = 2x - y + 3, \quad y^{ * } = - 2x + 2y - 6$

$x^{ * } = 4x + 3y - 1, \quad y^{ * } = - 3x - 2y + 1$

$x^{ * } = x, \quad y^{ * } = y$.

$x^{ * } = 2x + 3y, \quad y^{ * } = - y$

$x^{ * } = - x + y, \quad y^{ * } = x - y$

$x^{ * } = y - 9, \quad y^{ * } = 9x + 1$

$x^{ * } = y, \quad y^{ * } = - x + 1$

$x^{ * } = 2x + y - 3, \quad y^{ * } = 2x + 3y - 6$

$x^{ * } = 5x + 3y + 1, \quad y^{ * } = - 3x - y$

$x^{ * } = 3x - 2y + 5, \quad y^{ * } = 2x - y + 5$.

$x = x^{\prime } + y^{\prime } - 2, \quad y = 2x^{\prime } - y^{\prime } + 3$

$x = 3x^{\prime } - 4y^{\prime } - 5, \quad y = 4x^{\prime } + 3y^{\prime } + 1$.

taking the lines $x - y + 1 = 0, \quad x + y - 1 = 0$ respectively to the lines $3x + 2y - 3 = 0, 2x + 3y + 1 = 0$, and the point $A(1, 1)$ to the point $B(- 1, - 2)$

taking the lines $A_{1} x + B_{1} y + C_{1} = 0$ and $A_{2} x + B_{2} y + C_{2} = 0$ respectively to the $y$-axis and the $x$-axis, and the point $A\left(x_{0}, y_{0}\right)$ to the point $B(1, 1)$ (the point $\mathrm{A}$ does not lie on the given lines).

$x^{2} + y^{2} = 1$

$x^{2} - y^{2} = 1$

$xy = 2$

$y^{2} = - 6x$

$(3x + 4y - 1)(4x - 3y + 1) = 0$

$(x + y - 1)(x + y + 1) = 2$.

$x^{2} + y^{2} = 4$

$\frac{x^{2}}{4} - y^{2} = 1$

$(y + 1)^{2} = 8(x - 1)$

$(x + y - 1)(x - y - 1) = 1$

$(x + 2y - 2)(x + 2y + 2) = 0$.

Write the formulas of an affine transformation of the first kind taking the ellipse $\frac{x^{2}}{4} + y^{2} = 1$ to itself so that the point $A(1, \sqrt{3}/2)$ goes to the point $B(- 2, 0)$.

Solve the same problem for an affine transformation of the second kind.

the point $A_{1}(2, 1)$ goes to the point $B_{1}(4, 4)$, and the point $A_{2}(1$, $1 / 4)$ goes to the point $B_{2}(3, 9 / 4)$

the determinant of the transformation equals 1.