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Question 21 of 25

What is an equation of the line that is perpendicular to [tex]y-4=2(x-6)[/tex] and passes through the point [tex](-3,-5)[/tex]?

A. [tex]y+5=-\frac{1}{2}(x+3)[/tex]
B. [tex]y-5=-2(x-3)[/tex]
C. [tex]y-5=\frac{1}{2}(x-3)[/tex]
D. [tex]y+5=2(x+3)[/tex]


Sagot :

To find an equation of a line that is perpendicular to the given line [tex]\( y - 4 = 2(x - 6) \)[/tex] and passes through the point [tex]\((-3, -5)\)[/tex], let's follow the steps systematically.

1. Convert the given equation to slope-intercept form:

Start with the given equation:
[tex]\[ y - 4 = 2(x - 6) \][/tex]

Distribute the 2 on the right-hand side:
[tex]\[ y - 4 = 2x - 12 \][/tex]

Add 4 to both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = 2x - 8 \][/tex]

This is now in the slope-intercept form [tex]\( y = mx + b \)[/tex], where the slope ([tex]\( m \)[/tex]) is 2.

2. Find the slope of the perpendicular line:

Perpendicular lines have slopes that are negative reciprocals of each other. Therefore, the slope of the line perpendicular to the one given will be:
[tex]\[ m = -\frac{1}{2} \][/tex]

3. Use the point-slope form to find the new equation:

The point-slope form of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1) \)[/tex] is a point on the line.

Substitute the point [tex]\((-3, -5) \)[/tex] and the slope [tex]\( -\frac{1}{2} \)[/tex]:
[tex]\[ y - (-5) = -\frac{1}{2}(x - (-3)) \][/tex]

Simplify the equation:
[tex]\[ y + 5 = -\frac{1}{2}(x + 3) \][/tex]

4. Verification:

The expression [tex]\[ y + 5 = -\frac{1}{2}(x + 3) \][/tex] matches the form of option A.

Therefore, the equation of the line that is perpendicular to [tex]\( y - 4 = 2(x - 6) \)[/tex] and passes through the point [tex]\((-3, -5)\)[/tex] is:

[tex]\[ \boxed{y + 5 = -\frac{1}{2}(x + 3)} \][/tex]