The value of $\lambda$, if the lines $(2x+3y+4)+\lambda(6x-y+12)=0$ are — match Column $C_1$ with Column $C_2$. Column $C_1$: (a) parallel to $y$-axis is; (b) perpendicular to $7x+y-4=0$ is; (c) passes through $(1,2)$ is; (d) parallel to $x$ axis is. Column $C_2$: (i) $\lambda=-\dfrac{3}{4}$; (ii) $\lambda=-\dfrac{1}{3}$; (iii) $\lambda=-\dfrac{17}{41}$; (iv) $\lambda=3$.

The value of λ, if the lines (2x+3y+4)+λ(6x-y+12)= – 1st PUC Mathematics

NCERT Exemplar · Q58

\lambda

(a) parallel to

y

-axis is;

(b) perpendicular to

7x+y-4=0

is;

(1,2)

is;

(d) parallel to

x

axis is. Column

C_2

(i)

\lambda=-\dfrac{3}{4}

;

(ii)

\lambda=-\dfrac{1}{3}

;

(iii)

\lambda=-\dfrac{17}{41}

;

(iv)

\lambda=3

Karnataka PUCShort· 2mImportance★★★★★est

99% · 144/145 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Slope Calculation

Slope Calculation — From Intuition to Precision

Imagine you're walking up a hill. Some hills are gentle — you barely notice the climb. Others are so steep you have to lean forward and use your hands. That "steepness" is what slope measures. In mathematics, slope tells us how fast a line rises or falls as we move from left to right.

The Intuition: Rise Over Run

Take any two points on a straight line. As you walk from the left point to the right point, two things happen:

You move horizontally — that's the run.
You move vertically — that's the rise (upwards) or fall (downwards).

Slope is simply the ratio:

Slope = (vertical change) ÷ (horizontal change)

If you climb 3 metres while walking 5 metres forward, the slope is $3/5 = 0.6$ . If you descend 2 metres while walking 4 metres forward, the slope is $-2/4 = -0.5$ — negative because you're going downhill.

The Precise Definition

Given two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ on a non-vertical line, the slope $m$ is:

m = \frac{y_2 - y_1}{x_2 - x_1}

The numerator is the rise (change in $y$ ), the denominator is the run (change in $x$ ). The order matters: subtract the first point's coordinates from the second's, consistently.

Watch out

Never divide by zero. If $x_2 = x_1$ , the line is vertical — slope is undefined (not zero, not infinite — just undefined).

What the Number Tells You

Slope value	What the line does
$m > 0$	Rises left to right (uphill)
$m < 0$	Falls left to right (downhill)
$m = 0$	Horizontal (flat)
$m$ undefined	Vertical (straight up/down)

The larger the absolute value $|m|$ , the steeper the line. A slope of $5$ is much steeper than a slope of $0.2$ .

A Worked Example

Find the slope of the line through $(1, 2)$ and $(4, 8)$ .

Step 1: Label the points. Let $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (4, 8)$ .

Step 2: Compute the rise: $y_2 - y_1 = 8 - 2 = 6$ .

Step 3: Compute the run: $x_2 - x_1 = 4 - 1 = 3$ .

Step 4: Divide: $m = \frac{6}{3} = 2$ .

The line rises 2 units vertically for every 1 unit it moves right. …

Start your 14-day free trial to unlock the full solution →

For the family $(2+6\lambda)x+(3-\lambda)y+(4+12\lambda)=0$ : (a) $\to$ (iv) $\lambda=3$ ; (b) $\to$ (iii) $\lambda=-\tfrac{17}{41}$ ; (c) $\to$ (i) $\lambda=-\tfrac34$ ; (d) $\to$ (ii) $\lambda=-\tfrac13$ .

Solution

Rewrite the family of lines as

(2+6\lambda)x+(3-\lambda)y+(4+12\lambda)=0,\qquad \text{slope }m=-\frac{2+6\lambda}{3-\lambda}.

(a) Parallel to the $y$ -axis (vertical line, so no $y$ -term): $3-\lambda=0\Rightarrow\lambda=3.$ $\to$ (iv)

(b) Perpendicular to $7x+y-4=0$ (its slope is $-7$ , so the required slope is $\tfrac17$ ):

$-\frac{2+6\lambda}{3-\lambda}=\frac17\Rightarrow -7(2+6\lambda)=3-\lambda\Rightarrow -14-42\lambda=3-\lambda\Rightarrow \lambda=-\frac{17}{41}.$ $\to$ (iii)

(c) Passes through $(1,2)$ : …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$L_1$
›Reveal solutionSolution

$L_1$ has slope $\tan 150^\circ=-\tfrac{1}{\sqrt3}$ ; the parallel line through $(0,-2)$ is $x+\sqrt3\,y+2\sqrt3=0$ — option (D).

Slope of $L_1$ . A line making $150^\circ$ with the positive $x$ -axis has slope
$m=\tan 150^\circ=-\frac{1}{\sqrt3}.$
(The point $(0,-6)$ only fixes $L_1$ ; it is not needed for $L_2$ .) …
Start your 14-day free trial to see the full solution →
KCET 2024Set A-11 markMCQ
$\dfrac{x-1}{-3} = \dfrac{y-2}{2k} = \dfrac{z-3}{2}$
›Reveal solutionSolution

Read the direction ratios off the symmetric form of each line and set their dot product to zero.

1. Extract the direction ratios. A line written in symmetric form
$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$
has direction ratios $\langle a, b, c \rangle$ (the denominators). So:
$L_1: \frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2} \;\;\Longrightarrow\;\; \vec{b_1} = \langle -3,\ 2k,\ 2 \rangle$ $L_2: \frac{x-1}{3k} = \frac{y-5}{1} = \frac{z-6}{-5} \;\;\Longrightarrow\;\; \vec{b_2} = \langle 3k,\ 1,\ -5 \rangle$
2. The perpendicularity condition. Two lines are mutually perpendicular exactly when the angle between their direction vectors is $90^\circ$ . Since
$\cos\theta = \frac{\vec{b_1}\cdot\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}, \qquad \theta = 90^\circ \Rightarrow \cos\theta = 0$
the condition reduces to the dot product being zero:
$a_1a_2 + b_1b_2 + c_1c_2 = 0$
(The points on the lines are irrelevant — perpendicularity of lines depends only on direction, not position.)

3. Substitute and solve.
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$ $-9k + 2k - 10 = 0$ $-7k - 10 = 0$ $-7k = 10 \quad\Longrightarrow\quad k = -\frac{10}{7}$
4. Verify by back-substitution. With $k = -\tfrac{10}{7}$ : …
Start your 14-day free trial to see the full solution →
COMEDK 2023Set 2023-E1 markMCQ
$\theta$
›Reveal solutionSolution
Since the slope is negative, theta must lie strictly between 90 and 180 degrees, i.e. theta is an obtuse angle.

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 …
Start your 14-day free trial to see the full solution →
COMEDK 2023Set 2023-M1 markMCQ
$60^{\circ}$
›Reveal solutionSolution

With the base slope $m_1=3$ and angle $60^\circ$ , solving $\sqrt3=\left|\frac{m-3}{1+3m}\right|$ gives the pair $\frac{3-\sqrt3}{1+3\sqrt3}$ and $\frac{3+\sqrt3}{1-3\sqrt3}$ .

The line $y-3x+18=0$ has slope $m_1=3$ . If a line of slope $m$ makes $60^\circ$ with it:
$\tan60^\circ=\left|\frac{m-m_1}{1+m m_1}\right|=\left|\frac{m-3}{1+3m}\right|=\sqrt3.$
Case $+\sqrt3$ : $m-3=\sqrt3(1+3m)\Rightarrow m-3\sqrt3\,m=3+\sqrt3\Rightarrow m(1-3\sqrt3)=3+\sqrt3$ , so

$m=\frac{3+\sqrt3}{1-3\sqrt3}.$ …
Start your 14-day free trial to see the full solution →
COMEDK 2022Set 20221 markMCQ
$^\circ$
›Reveal solutionSolution

So the slopes are 1/3 and −3.

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. …
Start your 14-day free trial to see the full solution →
KCET 2019Set A-11 markMCQ
$\sqrt{3} \operatorname{cosec} 20^\circ - \sec 20^\circ =$
›Reveal solutionSolution

Put both terms over the common denominator $\sin20^\circ\cos20^\circ$ ; the numerator collapses to a single sine via the compound-angle formula and the denominator via the double-angle formula, and the $\sin40^\circ$ cancels.

Step 1 — Write everything in sine and cosine and combine.
$\sqrt{3}\,\operatorname{cosec}20^\circ - \sec20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} = \frac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}$
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:
$\sqrt{3}\cos 20^\circ - \sin 20^\circ = 2\!\left(\frac{\sqrt{3}}{2}\cos 20^\circ - \frac{1}{2}\sin 20^\circ\right)$
Since $\cos 30^\circ = \dfrac{\sqrt3}{2}$ and $\sin 30^\circ = \dfrac12$ , this is exactly the expansion of $\cos(A+B)$ :
$= 2\left(\cos 30^\circ \cos 20^\circ - \sin 30^\circ \sin 20^\circ\right) = 2\cos(30^\circ + 20^\circ) = 2\cos 50^\circ$
and using $\cos 50^\circ = \sin 40^\circ$ (complementary angles),
$\text{numerator} = 2\sin 40^\circ$
Step 3 — Simplify the denominator using the double-angle formula. …
Start your 14-day free trial to see the full solution →

←Previous questionQ57 — Match the entries of Column C₁ with their appropriate answers given un…→Next questionQ59 — The equation of the line through the intersection of the lines 2x-3y=0…

Q.The value of λ\lambdaλ, if the lines (2x+3y+4)+λ(6x−y+12)=0(2x+3y+4)+\lambda(6x-y+12)=0(2x+3y+4)+λ(6x−y+12)=0 are — match Column C1C_1C1​ with Column C2C_2C2​. Column C1C_1C1​:

Solution

🎓Unlock everything free for 14 days

Q.The value of $\lambda$ , if the lines $(2x+3y+4)+\lambda(6x-y+12)=0$ are — match Column $C_1$ with Column $C_2$ . Column $C_1$ :

Unlock everything free for 14 days