Q.Find the values of $k$ for which the line $(k - 3)x - (4 - k^2)y + k^2 - 7k + 6 = 0$ is

The line $x+\sqrt3\,y=10$ has normal direction $(1,\sqrt3)$. Since $OP\perp$ line, $OP$ points along this normal.

The angle $\alpha$ between $OP$ and the $y$-axis (unit $(0,1)$) satisfies

so $\alpha=30^\circ$.

Angle with $y$-axis $=60^\circ\Rightarrow$ angle with $x$-axis $=30^\circ$, so slope $m=\tan30^\circ=\tfrac1{\sqrt3}$.

Line through $(0,5)$: $y=\tfrac1{\sqrt3}x+5\Rightarrow \sqrt3y=x+5\sqrt3\Rightarrow x-\sqrt3y+5\sqrt3=0$.

Check the $x$-intercept: $y=0\Rightarrow x=-5\sqrt3<0$ — it meets the $x$-axis in the negative direction, as required.

Write the line in normal form. The normal vector to $\sqrt3x+y-24=0$ is $(\sqrt3,1)$ with magnitude $\sqrt{3+1}=2$. Since the constant $24>0$, the perpendicular from the origin points along $(\tfrac{\sqrt3}{2},\tfrac12)$.

Thus $\cos\alpha=\tfrac{\sqrt3}{2}$ and $\sin\alpha=\tfrac12$, so $\alpha=30^\circ$.

The angle of inclination is $120^\circ$, so

Using slope-intercept form with intercept $c=-5$:

Rearrange:

The line joining $(1,0)$ and $(4,3)$ has slope

so it makes an angle of $45^\circ$ with the $x$-axis.

Rotating counter-clockwise through $15^\circ$ about $(1,0)$ makes the new angle $45^\circ+15^\circ=60^\circ$, giving slope $\tan 60^\circ=\sqrt{3}$.

The new line passes through $(1,0)$:

The equation is $x+\sqrt{3}y=2\sqrt{3}$. The magnitude of the coefficient vector is $\sqrt{1^2+(\sqrt3)^2}=2$.

Dividing throughout by $2$:

Comparing with $x\cos\alpha+y\sin\alpha=p$: $\cos\alpha=\dfrac12$, $\sin\alpha=\dfrac{\sqrt3}{2}$, so $\alpha=\dfrac{\pi}{3}$ and $p=\sqrt3$.

$y=2+\frac23(0-1)=2-\frac23=\frac43$.

Concept and Intuition

Reflection in the x-axis reverses the sign of the slope of the incident ray.

Step-by-Step Solution

1. Incident ray from $(0,1)$ to $(\sqrt3,0)$ has slope $\dfrac{0-1}{\sqrt3-0}=-\tfrac{1}{\sqrt3}$.
2. On reflection about the x-axis the slope changes sign.
3. Reflected slope $=+\tfrac{1}{\sqrt3}$.

Common Mistakes

• Reflecting about the y-axis or keeping the same slope.

ANSWER: A

Concept and Intuition

The slope is $\tan\alpha$; with $\cos\alpha=\tfrac{\sqrt3}{2}$, $\alpha=30^\circ$ and $\tan\alpha=\tfrac1{\sqrt3}$. Use the $y$-intercept $-2$.

Step-by-Step Solution

1. Slope $=\tan 30^\circ=\tfrac1{\sqrt3}$.
2. $y=\tfrac1{\sqrt3}x-2$.
3. Multiply by $\sqrt3$: $\sqrt3 y=x-2\sqrt3\Rightarrow \sqrt3 y-x+2\sqrt3=0$.

Common Mistakes

• Using $\tan\alpha=\sqrt3$ (that is for $\cos\alpha=\tfrac12$).

ANSWER: D