The problem provides the coordinates of three points $A(1,2)$, $B(4,2)$, and $C(4,5)$. A vector $n = (-1,2)$ is also given. The problem consists of several sub-problems: a. Find the reflection of triangle $ABC$ with respect to the line $y=x$. b. Find the equation of the line (D) passing through B and perpendicular to vector $n$. c. A circle (C) is given with center $I(-2,1)$ and radius $r=3$. Find the distance from the center $I$ of the circle (C) to the line (D). Does the line (D) intersect circle (C)? d. Find the lengths of $BA$, $CA$ and $BC$. What can be said about triangle $ABC$? e. Find the parametric equation of the line (L) passing through point A and parallel to the vector $n$.

1. Problem Description

The problem provides the coordinates of three points $A(1,2)$, $B(4,2)$, and $C(4,5)$. A vector $n = (-1,2)$ is also given. The problem consists of several sub-problems:

a. Find the reflection of triangle $ABC$ with respect to the line $y=x$.

b. Find the equation of the line (D) passing through B and perpendicular to vector $n$.

c. A circle (C) is given with center $I(-2,1)$ and radius $r=3$. Find the distance from the center $I$ of the circle (C) to the line (D). Does the line (D) intersect circle (C)?

d. Find the lengths of $BA$, $CA$ and $BC$. What can be said about triangle $ABC$?

e. Find the parametric equation of the line (L) passing through point A and parallel to the vector $n$.

2. Solution Steps

a. Reflection across $y=x$ swaps the $x$ and $y$ coordinates.

$A'(2,1)$, $B'(2,4)$, $C'(5,4)$

b. The line (D) passes through $B(4,2)$ and is perpendicular to $n=(-1,2)$. The equation of a line in the form $ax+by+c=0$ has a normal vector $(a,b)$. Therefore the line (D) has the equation $-x + 2y + c = 0$. Since $B(4,2)$ lies on (D), we have $-4+2(2)+c=0$, which implies $c=0$.

Thus the equation of the line (D) is $-x+2y=0$, or $x-2y=0$.

c. Distance from point $I(-2,1)$ to line $D: x-2y=0$ is given by the formula:

$d = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$. In our case, $x_0 = -2$, $y_0 = 1$, $a = 1$, $b = -2$, $c = 0$.

$d = \frac{|1(-2)-2(1)+0|}{\sqrt{1^2+(-2)^2}} = \frac{|-2-2|}{\sqrt{1+4}} = \frac{|-4|}{\sqrt{5}} = \frac{4}{\sqrt{5}} = \frac{4\sqrt{5}}{5} \approx 1.789$.

Since $d = \frac{4}{\sqrt{5}} < r = 3$, the line (D) intersects the circle (C).

d. $BA = \sqrt{(4-1)^2 + (2-2)^2} = \sqrt{3^2+0^2} = 3$.

$CA = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{3^2+3^2} = \sqrt{18} = 3\sqrt{2}$.

$BC = \sqrt{(4-4)^2 + (5-2)^2} = \sqrt{0^2+3^2} = 3$.

Since $BA=BC$, triangle $ABC$ is an isosceles triangle.

Also, $BA^2 + BC^2 = 3^2+3^2 = 9+9=18 = (3\sqrt{2})^2 = CA^2$. Therefore, triangle $ABC$ is a right triangle.

e. The line (L) passes through $A(1,2)$ and is parallel to $n=(-1,2)$.

Parametric equation:

$x = 1 - t$

$y = 2 + 2t$

3. Final Answer

a. $A'(2,1)$, $B'(2,4)$, $C'(5,4)$

b. $x-2y=0$

c. $d = \frac{4\sqrt{5}}{5}$. Yes, the line (D) intersects circle (C).

d. $BA = 3$, $CA = 3\sqrt{2}$, $BC = 3$. Triangle $ABC$ is an isosceles right triangle.

e. $x = 1 - t$, $y = 2 + 2t$