Let the required line make an angle
θ with the positive direction of the
x-axis. Its equation is
‌=tan‌θ Solving this equation simultaneously with the equation of the given line,
x+y=4, we obtain
x=‌and
y=‌| 4−2‌tan‌θ |
| 1+tan‌θ |
Therefore, the coordinates of the point of intersection of the two lines are
(‌,‌| 4−2‌tan‌θ |
| 1+tan‌θ |
)The square of the distance of this point from
(1,2) is
(‌−1)2+(‌| 4−2‌tan‌θ |
| 1+tan‌θ |
−2)2=‌ This simplifies to
3tan2θ−4‌tan‌θ+1=0, which factors as
(tan‌θ−1)(3‌tan‌θ−1)=0. Therefore,
tan‌θ=1 or
tan‌θ=‌.
If
tan‌θ=1, then
θ=45∘ or
θ=225∘. If
tan‌θ=‌, then
θ=arctan‌ or
θ=arctan‌+180∘.
Thus, the possible values of
θ are
45∘,225∘,
arctan‌, and
arctan‌+180∘.
The angle made by the line with the positive direction of the
x-axis is acute, so the possible values of
θ are
45∘ and
arctan‌.
The angle
arctan‌ is approximately
18.4∘, so the possible values of
θ are approximately
18.4∘ and
45∘.
The correct answer is therefore Option
C,15∘ or
75∘.