PRECALCULUS:...COMMON CORE ED.-W/ACCESS
PRECALCULUS:...COMMON CORE ED.-W/ACCESS
10th Edition
ISBN: 9780134781839
Author: Demana
Publisher: PEARSON
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Chapter P.4, Problem 70E
Solution

The equation of the line tangent to the semicircle at the point 3,4 .

  y−4=−34x−3

Given:

The semicircle of radius 5 centered at 0, 0 ,

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter P.4, Problem 70E

Concept Used:

  • The slope-intercept form of a line is y=mx+b, where m is the slope of the line and b is the y -intercept of the line.
  • Slope of the line passing through the points x1,y1 and x2,y2 is given by m=y2−y1x2−x1
  • The slopes of two perpendicular lines are negative reciprocal of each other.
  • The slopes of two parallel lines are equal to each other.

Calculation:

Note that the equation of radius connecting the center 0, 0 and the point on the circumference 3,4 can be obtained as

  y−y1=y2−y1x2−x1x−x1y−0=4−03−0x−0y=43x−0y=43x

Comparing with the slope-intercept form y=mx+b , the slope is m=43 .

Since the tangent is perpendicular to the radius, the slope of the tangent line will be negative reciprocal of the slope of the radius, that is, the slope of tangent line is m=−34 .

Now substitute m=−34 and x1,y1=3,4 into the point-slope form of line to get

  y−y1=mx−x1y−4=−34x−3

Thus, the equation of the tangent line is y−4=−34x−3 .

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Chapter P.4, Problem 69E
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Chapter P.4, Problem 71E

Chapter P Solutions

PRECALCULUS:...COMMON CORE ED.-W/ACCESS

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