Find the equation of a straight line that contains the point [tex]\((-3, -2)\)[/tex] and is parallel to the line: [tex]\( y = 2x + 3 \)[/tex].



Answer :

To find the equation of a line that is parallel to a given line and passes through a specific point, follow these steps:

1. Identify the slope of the given line:
The given line is [tex]\( y = 2x + 3 \)[/tex]. The slope of this line is the coefficient of [tex]\( x \)[/tex], which is [tex]\( 2 \)[/tex]. Since the new line is parallel to the given line, it will have the same slope. Therefore, the slope of the new line is also [tex]\( 2 \)[/tex].

2. Use the point-slope form of the equation:
We have the slope ([tex]\( m = 2 \)[/tex]) and the point ([tex]\(-3, -2\)[/tex]) through which the new line passes. The point-slope form of a line's equation is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]

3. Substitute the given point and the slope into the point-slope form equation:
Here, [tex]\( x_1 = -3 \)[/tex] and [tex]\( y_1 = -2 \)[/tex]. Substituting these values in, we get:
[tex]\[ y - (-2) = 2(x - (-3)) \][/tex]
Simplifying this, we obtain:
[tex]\[ y + 2 = 2(x + 3) \][/tex]

4. Expand and simplify to get the equation in slope-intercept form:
Now distribute the [tex]\( 2 \)[/tex] on the right-hand side:
[tex]\[ y + 2 = 2x + 6 \][/tex]

To get the equation in the form [tex]\( y = mx + c \)[/tex], isolate [tex]\( y \)[/tex] by subtracting [tex]\( 2 \)[/tex] from both sides:
[tex]\[ y = 2x + 6 - 2 \][/tex]

Simplifying further, we have:
[tex]\[ y = 2x + 4 \][/tex]

Hence, the equation of the straight line that is parallel to [tex]\( y = 2x + 3 \)[/tex] and passes through the point [tex]\((-3, -2)\)[/tex] is:
[tex]\[ y = 2x + 4 \][/tex]