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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], let's go through the necessary steps:
1. Determine the slope of the original line:
The given line has an equation [tex]\( y = \frac{3}{5}x + 10 \)[/tex]. The slope [tex]\( m \)[/tex] of this line is [tex]\( \frac{3}{5} \)[/tex].
2. Find the slope of the perpendicular line:
Lines that are perpendicular to each other have slopes that are negative reciprocals. Hence, the slope of the line perpendicular to [tex]\( \frac{3}{5} \)[/tex] is:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{\frac{3}{5}} = -\frac{5}{3} \][/tex]
3. Use the point-slope form of the equation of a line:
The point-slope form is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( (x_1, y_1) \)[/tex] is a point on the line and [tex]\( m \)[/tex] is the slope. Here, we need to substitute the point [tex]\((15, -5)\)[/tex] and the slope [tex]\(-\frac{5}{3}\)[/tex].
Start with the point-slope form:
[tex]\[ y - (-5) = -\frac{5}{3}(x - 15) \][/tex]
Simplify inside the parentheses:
[tex]\[ y + 5 = -\frac{5}{3}(x - 15) \][/tex]
4. Distribute the slope [tex]\(-\frac{5}{3}\)[/tex]:
[tex]\[ y + 5 = -\frac{5}{3}x + \left(-\frac{5}{3}\right) \cdot 15 \][/tex]
Calculate the constant term:
[tex]\[ -\frac{5}{3} \cdot 15 = -25 \][/tex]
So the equation becomes:
[tex]\[ y + 5 = -\frac{5}{3}x - 25 \][/tex]
5. Solve for [tex]\( y \)[/tex] to get the equation in slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y = -\frac{5}{3}x - 25 - 5 \][/tex]
Simplify the constant term:
[tex]\[ y = -\frac{5}{3}x - 30 \][/tex]
However, upon closer examination:
[tex]\[ y + 5 = -\frac{5}{3}x + 25 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{5}{3}x + 25 - 5 \][/tex]
[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]
The correct answer among the given choices is:
[tex]\[ \boxed{y = -\frac{5}{3}x + 20} \][/tex]
Hence, the correct option is:
[tex]\[ \boxed{4 \text{ (Corresponding to: } y=-\frac{5}{3} x+20 \text{)}} \][/tex]
1. Determine the slope of the original line:
The given line has an equation [tex]\( y = \frac{3}{5}x + 10 \)[/tex]. The slope [tex]\( m \)[/tex] of this line is [tex]\( \frac{3}{5} \)[/tex].
2. Find the slope of the perpendicular line:
Lines that are perpendicular to each other have slopes that are negative reciprocals. Hence, the slope of the line perpendicular to [tex]\( \frac{3}{5} \)[/tex] is:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{\frac{3}{5}} = -\frac{5}{3} \][/tex]
3. Use the point-slope form of the equation of a line:
The point-slope form is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( (x_1, y_1) \)[/tex] is a point on the line and [tex]\( m \)[/tex] is the slope. Here, we need to substitute the point [tex]\((15, -5)\)[/tex] and the slope [tex]\(-\frac{5}{3}\)[/tex].
Start with the point-slope form:
[tex]\[ y - (-5) = -\frac{5}{3}(x - 15) \][/tex]
Simplify inside the parentheses:
[tex]\[ y + 5 = -\frac{5}{3}(x - 15) \][/tex]
4. Distribute the slope [tex]\(-\frac{5}{3}\)[/tex]:
[tex]\[ y + 5 = -\frac{5}{3}x + \left(-\frac{5}{3}\right) \cdot 15 \][/tex]
Calculate the constant term:
[tex]\[ -\frac{5}{3} \cdot 15 = -25 \][/tex]
So the equation becomes:
[tex]\[ y + 5 = -\frac{5}{3}x - 25 \][/tex]
5. Solve for [tex]\( y \)[/tex] to get the equation in slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y = -\frac{5}{3}x - 25 - 5 \][/tex]
Simplify the constant term:
[tex]\[ y = -\frac{5}{3}x - 30 \][/tex]
However, upon closer examination:
[tex]\[ y + 5 = -\frac{5}{3}x + 25 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{5}{3}x + 25 - 5 \][/tex]
[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]
The correct answer among the given choices is:
[tex]\[ \boxed{y = -\frac{5}{3}x + 20} \][/tex]
Hence, the correct option is:
[tex]\[ \boxed{4 \text{ (Corresponding to: } y=-\frac{5}{3} x+20 \text{)}} \][/tex]
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