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Question

Find the equation of a straight line passing through (3, 4) and having slope 3.

The correct answer is
3x - y - 5 = 0

Finding the Equation of a Straight Line

This problem requires us to find the equation of a straight line given specific information: a point it passes through and its slope.

Understanding the Given Information

  • Point: The line passes through the coordinates $(3, 4)$. We denote this point as $(x_1, y_1)$, so $x_1 = 3$ and $y_1 = 4$.
  • Slope: The slope of the line is given as 3. We denote the slope as $m$, so $m = 3$.

Key Formula: The Point-Slope Form

The most direct way to find the equation of a line when you have a point and the slope is by using the point-slope form. The formula is:

$$y - y_1 = m(x - x_1)$$

Where:

  • $(x_1, y_1)$ are the coordinates of the known point.
  • $m$ is the slope of the line.
  • $(x, y)$ represents any point on the line.

Step-by-Step Calculation

Let's substitute the given values into the point-slope formula:

  1. Substitute the point $(3, 4)$ and slope $m = 3$:
    $$y - 4 = 3(x - 3)$$
  2. Distribute the slope on the right side:
    Multiply the 3 by both terms inside the parentheses $(x - 3)$.
    $$y - 4 = 3 \cdot x - 3 \cdot 3$$ $$y - 4 = 3x - 9$$
  3. Rearrange the equation into the standard form ($Ax + By + C = 0$):
    To get the standard form, we move all terms to one side of the equation. Subtract $y$ from both sides and add 4 to both sides.
    $$0 = 3x - y - 9 + 4$$ $$0 = 3x - y - 5$$ So, the equation of the straight line is $3x - y - 5 = 0$.

Matching the Correct Option

Now, we compare our derived equation, $3x - y - 5 = 0$, with the given options:

  • Option 1: $3x - y - 5 = 0$
  • Option 2: $3x - y + 9 = 0$
  • Option 3: $x - 3y - 9 = 0$
  • Option 4: $3x - y - 9 = 0$

Our calculated equation perfectly matches Option 1.

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Important Questions from General Equation of a Line

  1. Two straight lines passing through the point A(3, 2) cut the line 2y = x + 3 and x-axis perpendicularly at P and Q respectively. The equation of the line PQ is

  2. Lines x = ay + b, z = cy + d

    and x = a'y + b', z = c'y + d'

    are perpendicular, if

  3. Equation of the line perpendicular to x - 2y = 1 and passing through (1, 1) is:

  4. If (2, 1), (–1, –2), (3, 3) are the midpoints of the sides BC, CA, AB of a triangle ABC, then equation of the line BC is

  5. If P(3, 4) is the mid-point of a line segment between the axes, then what is the equation of the line?

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