The vertex of an equilateral triangle is (2,3) and the equation of the opposite side is x+y=2. Then the other

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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 two sides through vertex $(2,3)$ make $60°$ angles with the base $x+y=2$ ; rotating the base's slope by $\pm 60°$ gives slopes $2\pm\sqrt{3}$ , yielding the equations $y-3=(2\pm\sqrt{3})(x-2)$ .

Why this works: geometry meets slope rotation

An equilateral triangle has all angles equal to $60°$ . If one side lies along $x+y=2$ and the opposite vertex is at $(2,3)$ , the two sides meeting at that vertex must each make a $60°$ angle with the base. The problem reduces to finding lines through $(2,3)$ that are inclined at $60°$ to the given line.

The key tool is the angle-between-two-lines formula. If two lines have slopes $m_1$ and $m_2$ , the acute angle $\theta$ between them satisfies

\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|.

We'll use this to rotate the base's slope by $60°$ in both directions.

Step-by-step derivation

Find the slope of the base.

The line $x+y=2$ can be rewritten as $y=-x+2$ , so its slope is $m_1=-1$ .
Set up the angle condition.

Let $m$ be the slope of one of the unknown sides. Since this side makes a $60°$ angle with the base,

\tan 60° = \left|\frac{m-(-1)}{1+m(-1)}\right| = \left|\frac{m+1}{1-m}\right|.

We know $\tan 60° = \sqrt{3}$ , so

\sqrt{3} = \left|\frac{m+1}{1-m}\right|.

Solve for the two slopes.

The absolute value gives two cases:

Case 1: $\frac{m+1}{1-m} = \sqrt{3}$

Cross-multiply: $m+1 = \sqrt{3}(1-m) = \sqrt{3}-\sqrt{3}m$ .

Collect terms: $m+\sqrt{3}m = \sqrt{3}-1$ , so $m(1+\sqrt{3}) = \sqrt{3}-1$ .

Thus

m = \frac{\sqrt{3}-1}{1+\sqrt{3}} = \frac{(\sqrt{3}-1)(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} = \frac{(\sqrt{3}-1)(1-\sqrt{3})}{1-3} = \frac{(\sqrt{3}-1)(1-\sqrt{3})}{-2}.

Expand the numerator: $(\sqrt{3}-1)(1-\sqrt{3}) = \sqrt{3}-3-1+\sqrt{3} = 2\sqrt{3}-4$ .

So $m = \frac{2\sqrt{3}-4}{-2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$ .

Case 2: $\frac{m+1}{1-m} = -\sqrt{3}$

Cross-multiply: $m+1 = -\sqrt{3}(1-m) = -\sqrt{3}+\sqrt{3}m$ .

Collect terms: $m-\sqrt{3}m = -\sqrt{3}-1$ , so $m(1-\sqrt{3}) = -\sqrt{3}-1$ .

Thus …

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

KEAM 2026Set eng-2026-04184 marksMCQ
$O$
›Reveal solutionSolution

The foot of perpendicular lies along the line's normal $(1,\sqrt3)$ , which makes $30^\circ$ with the $y$ -axis.

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 …
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KEAM 2026Set eng-2026-04194 marksMCQ
$y$
›Reveal solutionSolution

The line makes $30^\circ$ with the $x$ -axis (slope $1/\sqrt3$ ), passes through $(0,5)$ , and cuts the negative $x$ -axis.

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

Line through $(0,5)$ : $y=\tfrac1{\sqrt3}x+5\Rightarrow \sqrt3y=x+5\sqrt3\Rightarrow x-\sqrt3y+5\sqrt3=0$ . …
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KEAM 2026Set eng-2026-04194 marksMCQ
$\sqrt{3}x + y - 24 = 0$
›Reveal solutionSolution

The foot-of-perpendicular direction is the line's normal $(\sqrt3,1)$ , giving $\cos\alpha=\tfrac{\sqrt3}{2}$ , $\sin\alpha=\tfrac12$ . …

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KEAM 2026Set eng-2026-04204 marksMCQ
$y$
›Reveal solutionSolution

The line makes $120^\circ$ with the x-axis, so slope $m=\tan120^\circ=-\sqrt3$ . With y-intercept $-5$ , its equation $y=-\sqrt3x-5$ rearranges to $\sqrt3x+y+5=0$ .

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

$m=\tan120^\circ=-\sqrt3.$ …
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KEAM 2025Set eng-2025-04284 marksMCQ
$(1, 0)$
›Reveal solutionSolution

The line inclined at $45^\circ$ becomes inclined at $60^\circ$ after a $15^\circ$ rotation; its equation through $(1,0)$ is $\sqrt{3}x-y-\sqrt{3}=0$ .

The line joining $(1,0)$ and $(4,3)$ has slope
$m=\frac{3-0}{4-1}=1,$
so it makes an angle of $45^\circ$ with the $x$ -axis. …
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KEAM 2025Set eng-2025-04284 marksMCQ
$x + \sqrt{3}y = 2\sqrt{3}$
›Reveal solutionSolution

Normalize by the coefficient magnitude $2$ ; the cosine/sine give $\alpha=\tfrac{\pi}{3}$ and $p=\sqrt3$ .

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 $2$ :

$\frac{1}{2}x+\frac{\sqrt3}{2}y=\sqrt3.$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$\left(\dfrac{-1}{2}, 1\right)$
›Reveal solutionSolution

Slope through $\left(-\frac12,1\right)$ and $(1,2)$ is $\frac{2-1}{1+1/2}=\frac23$ ; setting $x=0$ in $y-2=\frac23(x-1)$ gives $y=\frac43$ . …

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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$(0,1)$
›Reveal solutionSolution
The reflected ray has slope $\tfrac{1}{\sqrt3}$ .

Concept and Intuition

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

Step-by-Step Solution

Incident ray from $(0,1)$ to $(\sqrt3,0)$ has slope $\dfrac{0-1}{\sqrt3-0}=-\tfrac{1}{\sqrt3}$ .

On reflection about the x-axis the slope changes sign. …
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KEAM 2021Set eng-2021-P2-B14 marksMCQ
$\alpha$
›Reveal solutionSolution
The line is $\sqrt3\,y-x+2\sqrt3=0$ .

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

Slope $=\tan 30^\circ=\tfrac1{\sqrt3}$ .

$y=\tfrac1{\sqrt3}x-2$ . …
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Q.The vertex of an equilateral triangle is (2,3)(2,3)(2,3) and the equation of the opposite side is x+y=2x+y=2x+y=2. Then the other two sides are y−3=(2±3)(x−2)y-3=(2\pm\sqrt{3})(x-2)y−3=(2±3​)(x−2).

Why this works: geometry meets slope rotation

Step-by-step derivation

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Q.The vertex of an equilateral triangle is $(2,3)$ and the equation of the opposite side is $x+y=2$ . Then the other two sides are $y-3=(2\pm\sqrt{3})(x-2)$ .

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