Parallel: same slope. Perpendicular: m₁m₂ = -1.

Find Equation of Line Parallel and Perpendicular to Given Line

Question

A line has equation 3x - 4y + 8 = 0. Find the equation of a line that: (a) is parallel to this line and passes through (2, -1) (b) is perpendicular to this line and passes through (2, -1)

Solution — Step by Step

Rewrite 3x - 4y + 8 = 0 in slope-intercept form:

4y = 3x + 8 \implies y = \frac{3}{4}x + 2

So the slope of the given line is $m = \dfrac{3}{4}$ .

For the parallel line: slopes are equal, so $m_{\parallel} = \dfrac{3}{4}$ .

For the perpendicular line: we need $m_1 \cdot m_2 = -1$ , so:

m_{\perp} = \frac{-1}{3/4} = -\frac{4}{3}

The perpendicular slope is the negative reciprocal — flip and negate.

We have slope $\dfrac{3}{4}$ passing through $(2, -1)$ :

y - (-1) = \frac{3}{4}(x - 2)

4(y + 1) = 3(x - 2)

4y + 4 = 3x - 6

\boxed{3x - 4y - 10 = 0}

Slope $-\dfrac{4}{3}$ through $(2, -1)$ :

y + 1 = -\frac{4}{3}(x - 2)

3(y + 1) = -4(x - 2)

3y + 3 = -4x + 8

\boxed{4x + 3y - 5 = 0}

Why This Works

The slope relationship is the heart of this problem. Two lines are parallel when they make the same angle with the x-axis — which means identical slopes. They are perpendicular when they meet at 90°, and trigonometry gives us the condition $m_1 \cdot m_2 = -1$ for this.

Notice the parallel line 3x - 4y - 10 = 0 and the original 3x - 4y + 8 = 0 differ only in the constant term. This is always true — parallel lines in the form ax + by + c = 0 have the same a and b coefficients, just a different c. This is a useful pattern to recognise quickly in MCQs.

For perpendicular lines, the coefficient swap happens because we invert and negate the slope. In the standard form ax + by + c = 0, a line perpendicular to it takes the form bx - ay + k = 0. Here, our original had 3x - 4y, and the perpendicular gives 4x + 3y — exactly this pattern.

Alternative Method

Direct formula for parallel/perpendicular lines in standard form.

Given line: ax + by + c = 0 and point $(x_1, y_1)$ .

Parallel line: $a(x - x_1) + b(y - y_1) = 0$
Perpendicular line: $b(x - x_1) - a(y - y_1) = 0$

For our problem, $a = 3$ , $b = -4$ , $(x_1, y_1) = (2, -1)$ :

Parallel:

3(x - 2) + (-4)(y + 1) = 0 \implies 3x - 4y - 10 = 0 \checkmark

Perpendicular:

(-4)(x - 2) - 3(y + 1) = 0 \implies -4x - 3y + 5 = 0 \implies 4x + 3y - 5 = 0 \checkmark

This method skips computing the slope entirely. In a CBSE board exam with 3 marks and a time crunch, this saves 30-40 seconds per question.

Memorise this: for line ax + by + c = 0, the perpendicular line through a point swaps a and b and flips one sign — giving bx - ay + k = 0. You can write the structure directly and just find k from the given point.

Common Mistake

The most frequent error: using $m_\perp = \dfrac{1}{m}$ instead of $m_\perp = \dfrac{-1}{m}$ .

Students flip the slope but forget to negate it. If $m = \dfrac{3}{4}$ , then $\dfrac{1}{m} = \dfrac{4}{3}$ — that gives you a line with a positive slope, which cannot be perpendicular to a line with a positive slope (think about it geometrically — they’d both tilt the same way).

The correct perpendicular slope is $-\dfrac{4}{3}$ . Always: flip AND negate.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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