Select the correct answer. Which equation represents the line that is perpendicular to [tex]$y=\frac{3}{4} x+1$[/tex] and passes through (-5,11)? A. [tex]$y=-\frac{4}{3} x+\frac{13}{3}$[/tex] B. [tex]$y=-\frac{4}{3} x+\frac{29}{3}$[/tex] C. [tex]$y=\frac{3}{4} x+\frac{59}{4}$[/tex] D. [tex]$y=\frac{3}{4} x-\frac{53}{4}$[/tex]
Answer (2)
The equation of the line perpendicular to y = 4 3 x + 1 and passing through the point ( − 5 , 11 ) is y = − 3 4 x + 3 13 , which matches option A.
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Find the slope of the perpendicular line: m = − 3 4 .
Use the point-slope form: y − 11 = − 3 4 ( x + 5 ) .
Simplify to slope-intercept form: y = − 3 4 x + 3 13 .
The equation of the line is y = − 3 4 x + 3 13 .
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. The given line is y = 4 3 x + 1 , and the point is ( − 5 , 11 ) .
Finding the Perpendicular Slope First, we need to find the slope of the line perpendicular to the given line. The slope of the given line is 4 3 . The slope of a line perpendicular to this line is the negative reciprocal of 4 3 , which is − 3 4 .
Applying Point-Slope Form Now we 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. In our case, m = − 3 4 and ( x 1 , y 1 ) = ( − 5 , 11 ) . Plugging these values into the point-slope form, we get:
y − 11 = − 3 4 ( x − ( − 5 ))
y − 11 = − 3 4 ( x + 5 )
Simplifying to Slope-Intercept Form Next, we simplify the equation to slope-intercept form ( y = m x + b ):
y − 11 = − 3 4 x − 3 4 ( 5 )
y − 11 = − 3 4 x − 3 20
y = − 3 4 x − 3 20 + 11
To add the numbers, we need a common denominator. Since 11 = 3 33 , we have:
y = − 3 4 x − 3 20 + 3 33
y = − 3 4 x + 3 13
Finding the Correct Option The equation of the line is y = − 3 4 x + 3 13 . Comparing this to the given options, we see that it matches option A.
Final Answer Therefore, the correct answer is A. y = − 3 4 x + 3 13
Examples
Understanding perpendicular lines is crucial in various real-world applications, such as architecture and navigation. For example, architects use perpendicular lines to ensure that walls are built at right angles, providing structural stability to buildings. Similarly, in navigation, understanding perpendicular relationships helps determine the shortest distance between two points, which is essential for efficient route planning. By mastering the concepts of slope and perpendicular lines, students can apply these principles to solve practical problems in diverse fields.
