PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
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Expert Solution & Answer
Chapter 6.2, Problem 40E
Solution
a.
The x -intercept and y -intercept of the line −2x+5y=10
The x -intercept A is (−5,0) and the y -intercept B is (0,2) .
Given:
The equation of the line is −2x+5y=10 .
Calculation:
The given equation is:
−2x+5y=10
The x -intercept will be the point where y=0 :
−2x+5(0)=10−2x=10x=−5
The y -intercept will be the point where x=0 :
−2(0)+5y=105y=10y=2
Hence, the x -intercept A is (−5,0) and the y -intercept B is (0,2) .
b.
Find the coordinates of P .
The coordinates of P are (−4.62,−0.95) and (−5.37,0.92) .
Given:
The equation of the line is −2x+5y=10 .
→AP is perpendicular to the line −2x+5y=10 .
|→AP|=1
Concept Used:
The relationship between the two lines with slopes m1 and m2 is:
m1×m2=−1
Calculation:
Let the coordinate P is (x0,y0) .
The coordinate of the point A is (−5,0)
The slope of the line →AP is:
y0−0x0+5y0x0+5
The slope of line −2x+5y=10 is:
−2x+5y=105y=2x+10y=25x+2
The slope of line −2x+5y=10 is 25 .
The line →AP and −2x+5y=10 are perpendicular. Therefore,
(y0x0+5)(25)=−1y0=−52(x0+5) ...(i)
The modulus of →AP given is 1. Therefore,
√(x0+5)2+(y0−0)2=1(x0+5)2+(y0−0)2=12(x0+5)2+y02=1
Substitute the value of y0 from equation (i) :
(x0+5)2+(−52(x0+5))2=1(x0+5)2+254(x0+5)2=1(x0+5)2(1+254)=1(x0+5)2=429x0+5=±2√29x0=−5±2√29x0=−5+2√29 and x0=−5−2√29x0≈−4.62 and x0≈−5.37
The value of y0 for x0=−4.62 is:
y0=−52(−4.62+5)y0=−0.95
The value of y0 for x0=−5.37 is:
y0=−52(−5.37+5)y0≈0.92
Hence, the coordinates of P are (−4.62,−0.95) and (−5.37,0.92) .
Chapter 6 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
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