# The points P and Q in the coordinate plane. ### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071 ### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1, Problem 17T

(a)

To determine

## To sketch: The points P and Q in the coordinate plane.

Expert Solution

### Explanation of Solution

The given points is P(3,1) and Q(5,6) .

In point P(3,1) , x-coordinate is negative and y-coordinate is positive, so it is lies on II Quadrant and point Q(5,6) both coordinates are positive so it is lies in I Quadrant.

The coordinate plane of two given points is shown below, Figure (1)

Figure (1) shows the points P(3,1) and Q(5,6) in coordinate plane.

(b)

To determine

### To find: The distance between P(−3,1) and Q(5,6) .

Expert Solution

The distance between P and Q is 89 .

### Explanation of Solution

Given:

The points is P(3,1) and Q(5,6) .

Calculation:

The distance formula of two points A(x1,y1) and B(x2,y2) is,

d(A,B)=(x2x1)2+(y2y1)2

Where,

d is the distance between two points.

x1 , x2 , y1 and y2 are coordinates of two points.

Substitute 5 for x2 , 3 for x1 , 6 for y2 and 1 for y1 from given points to the above formula to get distance,

d(P,Q)=(5(3))2+(61)2=(8)2+(5)2=64+25=89

Thus, the distance between P and Q is 89 .

(c)

To determine

### To find: The midpoint of line segment PQ.

Expert Solution

The midpoint of segment PQ is (1,72) .

### Explanation of Solution

Given:

The points is P(3,1) and Q(5,6) .

Calculation:

The formula of line segment from A(x1,y1) to B(x2,y2) is (x1+x22,y1+y22) .

Substitute 5 for x2 , 3 for x1 , 6 for y2 and 1 for y1 from given points in the above formula to get midpoint,

(x1+x22,y1+y22)=((3)+52,1+62)=(22,72)=(1,72)

Thus, the midpoint of segment PQ is (1,72) .

(d)

To determine

### To find: The slope of line passes through two pints P and Q.

Expert Solution

The slope of line PQ is 58 .

### Explanation of Solution

Given:

The points is P(3,1) and Q(5,6) .

Calculation:

The formula of slope of line passes through two points is,

m=y2y1x2x1

In above formula m is slope of line and (x1,y1) , (x2,y2) are coordinates of two points.

Substitute 6 for y2 , 1 for y1 , 5 for x2 , and 3 for x1 from given points to the above formula to get slope,

m=615(3)m=58

Thus, the slope of line PQ is 58 .

(e)

To determine

### To find: The perpendicular bisector of line PQ.

Expert Solution

The equation of perpendicular bisector is y=85x+5110 .

### Explanation of Solution

Given:

The points is P(3,1) and Q(5,6) .

Calculation:

The slope of line passes through points P(3,1) and Q(5,6) is m=58 that obtained from part (d).

The lines are perpendicular if,

m1m2=1

Where,

m1 slope of line passes through two points.

m2 is slope of perpendicular bisector.

Substitute 58 for m1 to get slope of perpendicular bisector,

58m2=1m2=158m2=85

The slope m2=85 is slope of perpendicular bisector of line PQ and perpendicular bisector is a line passes midpoint (1,72) of line segment PQ obtained from part (c) with slope 85 .

The formula of point-slope form of line is,

yy1=m(xx1)

Substitute 72 for y1 , 1 for x1 and 85 for min the above formula to get equation of perpendicular bisector,

y72=85(x1)y=85x+85+72y=85x+16+3510y=85x+5110

Thus, the equation of perpendicular bisector is y=85x+5110 .

(f)

To determine

### To find: The equation of circle with diameter PQ.

Expert Solution

The equation of circle is (x1)2+(y72)2=894 .

### Explanation of Solution

Given:

The points is P(3,1) , Q(5,6) and the line segment PQ is a diameter of circle.

Calculation:

The equation of circle with center (h,k) and radius r is,

(xh)2+(yk)2=r2 . (1)

The distance between points P(3,1) and Q(5,6) is 89 from part (b) that is a diameter of circle through given conditions.

The radius of circle is half of diameter that is,

r=892

The line segment PQ is a diameter of circle and the midpoint of line segment PQ is center of circle that is (1,72) obtained from part (c).

Substitute 1 for h, 72 for k and 892 for r in equation (1) to get the equation of circle,

(x1)2+(y72)2=(892)2(x1)2+(y72)2=894

Thus, the equation of circle is (x1)2+(y72)2=894 .

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