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

Slope of $L_1$. A line making $150^\circ$ with the positive $x$-axis has slope

(The point $(0,-6)$ only fixes $L_1$; it is not needed for $L_2$.)

Line $L_2$. $L_2$ is parallel to $L_1$, so it has the same slope $m=-\tfrac{1}{\sqrt3}$, and it crosses the $y$-axis $2$ units below the origin, i.e. its $y$-intercept is $-2$. Hence

Clear the surd. Multiply through by $\sqrt3$:

1. Extract the direction ratios. A line written in symmetric form

has direction ratios $\langle a, b, c \rangle$ (the denominators). So:

2. The perpendicularity condition. Two lines are mutually perpendicular exactly when the angle between their direction vectors is $90^\circ$. Since

the condition reduces to the dot product being zero:

(The points on the lines are irrelevant — perpendicularity of lines depends only on direction, not position.)

3. Substitute and solve.

4. Verify by back-substitution. With $k = -\tfrac{10}{7}$:

5. Note the distractors. $-\tfrac{7}{10}$ is the reciprocal (from writing $10k = -7$); $-10$ and $-7$ come from dropping the coefficient $-7$ or the constant $10$.

ANSWER: A

Concept: slope m = tan(theta), where theta is the inclination measured anticlockwise from the positive x-axis, 0 <= theta < 180 degrees.

On 0 <= theta < 180:

• tan theta > 0 for 0 < theta < 90 (acute) -> positive slope
• tan theta = 0 for theta = 0 -> line parallel to / coincident with the x-axis
• tan theta < 0 for 90 < theta < 180 (obtuse) -> negative slope

Since the slope is negative, theta must lie strictly between 90 and 180 degrees, i.e. theta is an obtuse angle.

ANSWER: A

The line $y-3x+18=0$ has slope $m_1=3$. If a line of slope $m$ makes $60^\circ$ with it:

Case $+\sqrt3$: $m-3=\sqrt3(1+3m)\Rightarrow m-3\sqrt3\,m=3+\sqrt3\Rightarrow m(1-3\sqrt3)=3+\sqrt3$, so

Case $-\sqrt3$: $m-3=-\sqrt3(1+3m)\Rightarrow m+3\sqrt3\,m=3-\sqrt3\Rightarrow m(1+3\sqrt3)=3-\sqrt3$, so

These match option (B).

Concept: tanθ = |(m − m₁)/(1 + m m₁)|.

Given line: 2x − y = −7 → y = 2x + 7 → m₁ = 2. Required angle θ = 45° → tanθ = 1.

|(m − 2)/(1 + 2m)| = 1

Case 1: (m − 2) = (1 + 2m) → −m = 3 → m = −3.

Case 2: (m − 2) = −(1 + 2m) → m − 2 = −1 − 2m → 3m = 1 → m = 1/3.

So the slopes are 1/3 and −3.

ANSWER: A

Step 1 — Write everything in sine and cosine and combine.

Step 2 — Simplify the numerator using the $a\cos\theta - b\sin\theta$ trick.

Factor out $2$ so the coefficients become a cosine and a sine of a standard angle:

Since $\cos 30^\circ = \dfrac{\sqrt3}{2}$ and $\sin 30^\circ = \dfrac12$, this is exactly the expansion of $\cos(A+B)$:

and using $\cos 50^\circ = \sin 40^\circ$ (complementary angles),

Step 3 — Simplify the denominator using the double-angle formula.

From $\sin 2\theta = 2\sin\theta\cos\theta$ with $\theta = 20^\circ$:

Step 4 — Divide.

The $\sin40^\circ$ cancels exactly, which is why the answer is a clean integer.

Step 5 — Numerical check. $\sqrt3/\sin20^\circ = 1.7321/0.3420 = 5.064$ and $1/\cos20^\circ = 1/0.9397 = 1.0642$. Difference $= 5.064 - 1.064 = 4.000$. ✓

ANSWER: C