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What is the equation of the line that is parallel to the line [tex]5x + 2y = 12[/tex] and passes through the point [tex]\((-2, 4)\)[/tex]?

A. [tex]y = -\frac{5}{2} x - 1[/tex]
B. [tex]y = -\frac{5}{2} x + 5[/tex]
C. [tex]y = \frac{2}{5} x - 1[/tex]
D. [tex]y = \frac{2}{5} x + 5[/tex]


Sagot :

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

1. Determine the slope of the given line: The equation of the given line is [tex]\(5x + 2y = 12\)[/tex]. First, we need to rewrite this equation in slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.

[tex]\[ 5x + 2y = 12 \][/tex]
Subtract [tex]\(5x\)[/tex] from both sides:
[tex]\[ 2y = -5x + 12 \][/tex]
Divide everything by 2 to solve for [tex]\(y\)[/tex]:
[tex]\[ y = \left(-\frac{5}{2}\right)x + 6 \][/tex]
So, the slope ([tex]\(m\)[/tex]) of the given line is [tex]\(-\frac{5}{2}\)[/tex].

2. Use the slope of the parallel line: Since parallel lines have the same slope, the slope of the line we are looking for will also be [tex]\(-\frac{5}{2}\)[/tex].

3. Use the point-slope form of the equation of a line: The point-slope form is given by [tex]\[y - y_1 = m(x - x_1)\][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line. Here, the point is [tex]\((-2, 4)\)[/tex], and the slope [tex]\(m\)[/tex] is [tex]\(-\frac{5}{2}\)[/tex].

4. Substitute the known values into the point-slope form equation:

[tex]\[ y - 4 = \left(-\frac{5}{2}\right)(x - (-2)) \][/tex]
Simplify inside the parentheses:
[tex]\[ y - 4 = \left(-\frac{5}{2}\right)(x + 2) \][/tex]
Distribute the slope [tex]\(-\frac{5}{2}\)[/tex]:
[tex]\[ y - 4 = \left(-\frac{5}{2}\right)x + \left(-\frac{5}{2}\right) \cdot 2 \][/tex]
Simplify the multiplication:
[tex]\[ y - 4 = -\frac{5}{2}x - 5 \][/tex]
Add 4 to both sides to solve for [tex]\(y\)[/tex]:
[tex]\[ y = -\frac{5}{2}x - 5 + 4 \][/tex]
Simplify the constants:
[tex]\[ y = -\frac{5}{2}x - 1 \][/tex]

Therefore, the equation of the line that is parallel to [tex]\(5x + 2y = 12\)[/tex] and passes through the point [tex]\((-2,4)\)[/tex] is [tex]\(\boxed{y = -\frac{5}{2}x - 1}\)[/tex].