What is the equation of the line that is perpendicular to the line [tex]$y=\frac{3}{5} x+10$[/tex] and passes through the point (15, -5)?
A. [tex]$y=\frac{3}{5} x-20$[/tex]
B. [tex]$y=-\frac{3}{5} x+20$[/tex]
C. [tex]$y=\frac{5}{3} x-20$[/tex]
D. [tex]$y=-\frac{5}{3} x+20$[/tex]
Answer (2)
Find the slope of the perpendicular line: The given line has a slope of 5 3 , so the perpendicular line has a slope of − 3 5 .
Use the point-slope form: Substitute the slope − 3 5 and the point ( 15 , − 5 ) into the point-slope form y − y 1 = m ( x − x 1 ) .
Simplify the equation: Simplify y − ( − 5 ) = − 3 5 ( x − 15 ) to get the slope-intercept form.
State the final equation: The equation of the line is y = − 3 5 x + 20 .
Explanation
Understanding the Problem The problem asks us to find the equation of a line that is perpendicular to a given line and passes through a specific point. We'll use the properties of perpendicular lines and the point-slope form to find the equation.
Finding the Perpendicular Slope The given line is y = 5 3 x + 10 . The slope of this line is 5 3 . A line perpendicular to this line will have a slope that is the negative reciprocal of 5 3 .
Calculating the Negative Reciprocal The negative reciprocal of 5 3 is − 3 5 . So, the slope of the line we are looking for is − 3 5 .
Using the Point-Slope Form We know the line passes through the point ( 15 , − 5 ) . We can use the point-slope form of a line, which is y − y 1 = m ( x − x 1 ) , where m is the slope and ( x 1 , y 1 ) is a point on the line.
Substituting Values Substituting m = − 3 5 and ( x 1 , y 1 ) = ( 15 , − 5 ) into the point-slope form, we get:
y − ( − 5 ) = − 3 5 ( x − 15 )
Simplifying the Equation Simplifying the equation:
y + 5 = − 3 5 x + 3 5 × 15
y + 5 = − 3 5 x + 25
y = − 3 5 x + 25 − 5
y = − 3 5 x + 20
Final Answer The equation of the line is y = − 3 5 x + 20 .
Examples
Understanding perpendicular lines is crucial in architecture and construction. For example, when designing a building, ensuring walls are perpendicular to the ground is essential for stability. If a wall's slope is represented by 5 3 , a supporting beam needs to be placed at a slope of − 3 5 to ensure it's perfectly perpendicular, providing maximum support. This principle extends to road design, bridge construction, and even interior design, where right angles and perpendicular lines are fundamental to creating functional and aesthetically pleasing spaces.
The equation of the line that is perpendicular to y = 5 3 x + 10 and passes through the point ( 15 , − 5 ) is y = − 3 5 x + 20 . This corresponds to option D. We derived this by using the negative reciprocal of the slope for perpendicular lines and employing the point-slope form of a line.
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