Equations of the lines through the point $(3,2)$ and making an angle of $45^\circ$ with the line $x-2y=3$ are ____.

Equations of the lines through the point (3,2) and making an angle of 45^° with the line x-2y=3 are __.

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(3,2)

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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. …

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The key idea is to use the slope formula for the angle between two lines. The slopes of the required lines are found by solving $\tan 45^\circ = \left| \frac{m - \frac12}{1 + \frac12 m} \right|$ , giving $m = 3$ and $m = -\frac13$ . The equations through $(3,2)$ are $3x - y = 7$ and $x + 3y = 9$ .

Concept and Intuition

When a problem says "a line makes an angle of $45^\circ$ with another line," it is always about the acute angle between their directions — that is, the angle between their slopes. The formula connecting the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is:

\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

Here, one line is given: $x - 2y = 3$ . Its slope is $m_1 = \frac12$ . We want lines through $(3,2)$ whose slope $m$ makes a $45^\circ$ angle with this line. So we set $\tan 45^\circ = 1$ and solve for $m$ .

Because the formula uses absolute value, we get two possible slopes — one steeper, one shallower — corresponding to the two lines that can be drawn through the point at exactly $45^\circ$ on either side of the given line.

Step-by-step solution

1. Find the slope of the given line.

Rewrite $x - 2y = 3$ as $y = \frac12 x - \frac32$ .

So the slope is $m_1 = \frac12$ .

2. Write the slope of the unknown line.

Let the required line through $(3,2)$ have slope $m$ . Its equation will be $y - 2 = m(x - 3)$ .

3. Apply the angle-between-lines formula.

We want $\theta = 45^\circ$ , so $\tan 45^\circ = 1$ .

Thus:

1 = \left| \frac{m - \frac12}{1 + \frac12 m} \right|

4. Remove the absolute value — two cases.

Case 1:

\frac{m - \frac12}{1 + \frac12 m} = 1

Multiply through: $m - \frac12 = 1 + \frac12 m$

$\Rightarrow m - \frac12 m = 1 + \frac12$

$\Rightarrow \frac12 m = \frac32$

$\Rightarrow m = 3$

Case 2:

\frac{m - \frac12}{1 + \frac12 m} = -1

Multiply: $m - \frac12 = -1 - \frac12 m$

$\Rightarrow m + \frac12 m = -1 + \frac12$

$\Rightarrow \frac32 m = -\frac12$

$\Rightarrow m = -\frac13$ …

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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 $L_2$ .) …
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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 $k = -\tfrac{10}{7}$ : …
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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 …
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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

$m=\frac{3+\sqrt3}{1-3\sqrt3}.$ …
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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. …
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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. …
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Q.Equations of the lines through the point (3,2)(3,2)(3,2) and making an angle of 45∘45^\circ45∘ with the line x−2y=3x-2y=3x−2y=3 are ____.

Concept and Intuition

Step-by-step solution

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Q.Equations of the lines through the point $(3,2)$ and making an angle of $45^\circ$ with the line $x-2y=3$ are ____.

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