Question

Find the equation of a straight line which passes through the point (2,-3) an d is parallel to the line 2x + y = 6

Answers

Answered by GPT-4o mini
To find the equation of a straight line that passes through the point \( (2, -3) \) and is parallel to the line given by the equation \( 2x + y = 6 \), we first need to determine the slope of the given line.

1. **Rewrite the line in slope-intercept form (y = mx + b):**
Start with the equation \( 2x + y = 6 \) and solve for \( y \):

\[
y = -2x + 6
\]

From this equation, we see that the slope \( m \) of the given line is \( -2 \).

2. **Use the slope for the new line:**
Since the new line is parallel to the original line, it will have the same slope of \( -2 \).

3. **Use the point-slope form of the line equation:**
The point-slope form of a line equation is given by:

\[
y - y_1 = m(x - x_1)
\]

Where \( (x_1, y_1) \) is the point the line passes through, and \( m \) is the slope. Here, \( (x_1, y_1) = (2, -3) \) and \( m = -2 \):

\[
y - (-3) = -2(x - 2)
\]

4. **Simplifying the equation:**

\[
y + 3 = -2(x - 2)
\]

Distributing the \( -2 \):

\[
y + 3 = -2x + 4
\]

Now, isolate \( y \):

\[
y = -2x + 4 - 3
\]

\[
y = -2x + 1
\]

5. **Final equation:**
The equation of the line that passes through the point \( (2, -3) \) and is parallel to the line \( 2x + y = 6 \) is:

\[
y = -2x + 1
\]
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