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If the line through $A=(4,-5)$ is inelined at an angle $45^{\circ}$ with the positive direction of the $X$-axis, then the coordinates of the two points on opposite sides of $A$ at a distance of $3 \sqrt{2}$ units are
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Verified Answer
The correct answer is:
$(7,-2),(1,-8)$
If the line is inclined at angle $45^{\circ}$ with the positive direction of the $X$-axis i.e. the points are situated at angle $\theta=45^{\circ}$ from $X$-axis at a distance $r=3 \sqrt{2}$ from point $(4,-5)$.
The two points can be given by
$\left(x_{0} \pm r \cos \theta, y \pm r \sin \theta\right)$
The points are
$$
\begin{aligned}
(4&\left.+3 \sqrt{2} \cos 45^{\circ},-5+3 \sqrt{2} \sin 45^{\circ}\right) \\
&=\left(4+3 \sqrt{2} \frac{1}{\sqrt{2}},-5+3 \sqrt{2} \times \frac{1}{\sqrt{2}}\right)=(7,-2)
\end{aligned}
$$
$$
\begin{aligned}
&\text { and }\left(4-3 \sqrt{2} \cos 45^{\circ},-5-3 \sqrt{2} \sin 45^{\circ}\right) \\
&=\left(4-3 \sqrt{2} \times \frac{1}{\sqrt{2}},-5-3 \sqrt{2} \times \frac{1}{\sqrt{2}}\right)=(1,-8)
\end{aligned}
$$
Thus, the points are $(1,-8)$ and $(7,-2)$.
The two points can be given by
$\left(x_{0} \pm r \cos \theta, y \pm r \sin \theta\right)$
The points are
$$
\begin{aligned}
(4&\left.+3 \sqrt{2} \cos 45^{\circ},-5+3 \sqrt{2} \sin 45^{\circ}\right) \\
&=\left(4+3 \sqrt{2} \frac{1}{\sqrt{2}},-5+3 \sqrt{2} \times \frac{1}{\sqrt{2}}\right)=(7,-2)
\end{aligned}
$$
$$
\begin{aligned}
&\text { and }\left(4-3 \sqrt{2} \cos 45^{\circ},-5-3 \sqrt{2} \sin 45^{\circ}\right) \\
&=\left(4-3 \sqrt{2} \times \frac{1}{\sqrt{2}},-5-3 \sqrt{2} \times \frac{1}{\sqrt{2}}\right)=(1,-8)
\end{aligned}
$$
Thus, the points are $(1,-8)$ and $(7,-2)$.
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