Question

If the line through $A=(4,-5)$ is inclined 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.

• A. $(7,2),(1,8)$
• B. $(7,2),(1,-8)$
• C. $(7,-2),(1,-8)$
• D. $(7,2),(-1,8)$

Answer 1

The correct option is C $(7,-2),(1,-8)$

If the line is inclined at angle ${45}^o$ with the positive direction of the x-axis, i.e. the points are situated at angles $\theta ={45}^o$ from x-axis at a distance $r=3\sqrt{2}$ from $(4,-5)$

The two points can be given by $(x_0\pm r\cos \theta ,y\pm r\sin \theta )$

The points are $(4+3\sqrt{2}\cos {45}^o,-5+3\sqrt{2}\sin {45}^o)=(4+3\sqrt{2}\times \frac{1}{\sqrt{2}},-5+3\sqrt{2}\times \frac{1}{\sqrt{2}})=(7,-2)$

and

$(4-3\sqrt{2}\cos {45}^o,-5-3\sqrt{2}\sin {45}^o)=(4-3\sqrt{2}\times \frac{1}{\sqrt{2}},-5-3\sqrt{2}\times \frac{1}{\sqrt{2}})=(1,-8)$

Thus, the points are $(1,-8)\&(7,-2)$