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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(2,3)

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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.

AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
$(1, 2)$
›Reveal solutionSolution

Slope $=3$ (parallel), point $(1,2)$ : $y-2=3(x-1)$ .

Parallel lines have equal slopes. The line $y = 3x - 1$ has slope $3$ .

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2025Set ANNUAL1 markMCQ
$-2$
(a) Both (A) and (R) are true and (R) is the correct explanation of (A)
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
›Reveal solutionSolution

Assertion is false, Reason is true — option (d).

Reason (R): The point-slope form $(y-y_1)=m(x-x_1)$ is a correct, standard formula. So (R) is true.

Assertion (A): “3 units to the left of origin” on the $x$ -axis is the point $(-3,0)$ . A line with slope $-2$ through it is

$y-0=-2(x-(-3))\Rightarrow y=-2x-6\Rightarrow 2x+y+6=0.$ …

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AHSEC Higher Secondary (HS) 1st Year Examination 2024Set ANNUAL1 mark
$x$
›Reveal solutionSolution

The slope of the $x$ -axis is $0$ .

The slope (gradient) of a line is defined as $m = \tan\theta$ , where $\theta$ is the angle the line makes with the positive direction of the $x$ -axis, measured counter-clockwise.

The $x$ -axis itself is inclined at $\theta = 0°$ to the positive $x$ -axis (it lies along it), so

$m = \tan 0° = 0$ …

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AHSEC Higher Secondary (HS) 1st Year Examination 2018Set ANNUAL1 mark
$y = x$
›Reveal solutionSolution

The inclination of $y=x$ is $45^\circ$ .

The line $y=x$ is of the form $y=mx+c$ with slope $m=1$ and $c=0$ . The inclination $\theta$ of a line satisfies $\tan\theta=m$ : …

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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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