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What is the equation of the line that is perpendicular to the line [tex][tex]$y=\frac{3}{5} x+10$[/tex][/tex] and passes through the point [tex][tex]$(15, -5)$[/tex][/tex]?

A. [tex][tex]$y=\frac{3}{5} x-20$[/tex][/tex]
B. [tex][tex]$y=-\frac{3}{5} x+20$[/tex][/tex]
C. [tex][tex]$y=\frac{5}{3} x-20$[/tex][/tex]
D. [tex][tex]$y=-\frac{5}{3} x+20$[/tex][/tex]

Sagot :

To find the equation of the line that is perpendicular to the line [tex]\(y = \frac{3}{5}x + 10\)[/tex] and passes through the point [tex]\((15, -5)\)[/tex], we need to follow these steps:

1. Determine the slope of the given line:

The given line is [tex]\(y = \frac{3}{5}x + 10\)[/tex]. The coefficient of [tex]\(x\)[/tex] is the slope of the line. So, the slope [tex]\(m\)[/tex] of the given line is:
[tex]\[ m = \frac{3}{5} \][/tex]

2. Find the slope of the perpendicular line:

For two lines to be perpendicular, the product of their slopes must be [tex]\(-1\)[/tex]. If [tex]\(m_1\)[/tex] is the slope of the first line, then the slope [tex]\(m_2\)[/tex] of the line perpendicular to it can be found using:
[tex]\[ m_2 = -\frac{1}{m_1} \][/tex]
Here, [tex]\(m_1 = \frac{3}{5}\)[/tex], so:
[tex]\[ m_2 = -\frac{1}{\frac{3}{5}} = -\frac{5}{3} \][/tex]
So, the slope of the perpendicular line is [tex]\(-\frac{5}{3}\)[/tex].

3. Use the point-slope form of the line equation:

The point-slope form of a line's equation is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
We have:
[tex]\[ (x_1, y_1) = (15, -5) \][/tex]
and the perpendicular slope [tex]\(m = -\frac{5}{3}\)[/tex]. Plugging these values in, we get:
[tex]\[ y - (-5) = -\frac{5}{3}(x - 15) \][/tex]
Simplifying, this becomes:
[tex]\[ y + 5 = -\frac{5}{3}(x - 15) \][/tex]

4. Simplify the equation to the slope-intercept form [tex]\(y = mx + b\)[/tex]:

Distribute [tex]\(-\frac{5}{3}\)[/tex] on the right-hand side:
[tex]\[ y + 5 = -\frac{5}{3}x + 25 \][/tex]
Subtract 5 from both sides:
[tex]\[ y = -\frac{5}{3}x + 20 \][/tex]

Therefore, the equation of the line that is perpendicular to [tex]\(y = \frac{3}{5}x + 10\)[/tex] and passes through the point [tex]\((15, -5)\)[/tex] is:
[tex]\[ y = -\frac{5}{3}x + 20 \][/tex]

From the given options, the correct answer is:
[tex]\[ y = -\frac{5}{3}x + 20 \][/tex]
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