Answer to Question #247163 in Analytic Geometry for beza

Question

Let L be the line given by the vector equation (x,y)=(1,1) + t( sqrt(3),1), t$\epsilon$R. What is the equation of the image of L after being rotated 15 deg about (1,1) and then translated by vector u=(-1,1)

Answer 1

The equation of a line in vector form is $a+\lambda b$

where a is the position vector of a point on the line and b is a vector parallel to the line.

Rotating the line through and angle $\theta$ is equivalent to rotating b through the same angle.

In this case: $b=(\sqrt 3,1)$

The slope of the line segment from the origin to the head of this vector is $1/\sqrt 3$

and hence its inclination to the positive X axis is $arctan(1/\sqrt 3)$

If we rotate this vector by $15\degree$its inclination to the positive X axis becomes

$arctan(1/\sqrt 3)+15\degree$

Hence the slope of b₁ after rotating:

$tan(arctan(1/\sqrt 3)+15\degree)=tan45\degree =1$

b₁ can then be represented by the position vector of point $(1,1)$

Translating the line through u=(-1,1) is equivalent to translating the given point on the line by

$(-1,1)$

Hence, after this translation the point (1,1) becomes (0,2)

Thus, it can be seen that after the given rotation and translation the vector equation of the line becomes

$(x,y)=(0,2)+t(1,1)$