analytic 1.pdf

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Excel Review Center Solutions to Take Home Exam Analytic Geometry 1

Cebu: JRT Bldg., Imus Avenue, Cebu City Tel. 2685989 90 | 09173239235 Manila: 3rd

& 4th

Fl. CMFFI Bldg. R. Papa St. Sampaloc Tel. 7365291

Problem 1

Solution:

( )( ) ( )2 2

3 3 2 4d

d 6

+ =

=

Problem 2

Solution:

( )

( )

m

m

m

4 8x 6

2

6 2y 2

2

P 6,2

+= =

+ = =

Problem 3

Solution:

( )

( )

m1

1 2

1 2

1 2

1 2

m2

2 3

2 3

2 3

2 3

P 2,5 :

x x2

2x x 4 I

y y5

2

y y 10 IV

P 4,2 :

x x4

2

x x 8 II

y y2

2

y y 4 V

+=

+ =

+=

+ =

+=

+ =

+=

+ =

( )m3

1 3

1 3

1 3

1 3

1 2

2 3

1 3

1

2

3

P 1,1 :

x x1

2

x x 2 III

y y1

2

y y 2 VI

From equations I,II & III:

x x 4 I

x x 8 II

x x 2 III

Using calculator:

x 1

x 5

x 3

+=

+ =

+=

+ =

+ =

+ =

+ =

=

=

=

1 2

2 3

1 3

1 2 3

From equations IV,V & VI:

y y 10 IV

y y 4 V

y y 2 VI

Using calculator:

y 4; y 6; y 2

+ =

+ =

+ =

= = =

( ) ( ) ( )

( ) ( ) ( )

1 1 2 2 3 3Vertices are x ,y , x ,y & x ,y :

1,4 , 5,6 & 3, 2

Problem 4

Solutiion:

( )4 1 3

m1 3 2

= =

Problem 5

Solution:

( ) ( )

( )

( ) ( )

( )

1

1

2

2

2 1

1 2

: 2,1 , 7,3m

3 1 2m

7 2 9

: 2,1 , 1,5m

5 1 4m

1 2 3

m mtan

1 m m

4 2

3 9tan4 2

13 9

40 36 '

= =

= =

=

+

=

+

=

Problem 6

Solution:

1

2

2x 3y 6 0

2m

3

4x y 3 0

4m 4

1

+ =

=

+ =

= =

( )

1 2

1 2

m mtan

1 m m

24

3tan2

1 43

14tan

5

=

+

=

+

=

Problem 7

Solution:

( ) ( )

( )

( ) ( )

( )

( )

( ) ( )

( )

1

2

1 3 y 2m

3 1 x 4

4 x 4 2 y 2

4x 16 2y 4

4x 2y 12 I

3 2 y 1m

1 4 x 3

1 x 3 5 y 1

x 5y 8 II

From I and II:Using calculator

4x 2y 12 I

x 5y 8 II

x 2

y 2

D(x,y) D 2,2

= =

+ = +

+ = +

=

= =

=

+ =

=

+ =

=

=

Problem 8

Solution:

( )( ) ( )

( ) ( )

m

m

P 2,1 midpointC 0, 2 &D x,y ends of diameter

x 0x 2

2

x 4

y 2y 1

2

2 y 2

y 4

D x,y D 4,4

+= =

=

= =

=

=

Problem 9

Solution:

The median passes through the point C andmidpoint of AB.

m

m

m

m

m

0 6x

2

x 3

0 0y

2

y 0Median is vertical line:

Thus, equation is x x

x 3

+=

=

+=

=

=

=

Problem 10

Solution:

The point of concurrency of the medians iscalled the centroid.

(-2,1)

(1,5)(10,7)

(7,3)

m1

m2

(x,y)(3,1)

(-4,-2)(1,-3)

m1

m1

m2

m2

A(0,0) B(6,0)

C(3,3)

(xm,ym)

M


2/5



& 4th


1 2 3c

c

1 2 3c

c

centroid

x x xx

3

1 2 5 8x

3 3

y y yy

3

2 4 0y 2

3

8P ,23

+ +=

+ += =

+ +=

+ += =

Problem 11

Solution:

( )( )

( )( )( )

( )

( )

Using Two-point form:

1 2y 2 x 1

5 1

3y 2 x 1

6

1y 2 x 1

2

=

+ = +

+ = +

Problem 12

Solution:

Slope-intercept form

1y x 2

3

3y x 6

x 3y 6

=

=

=

Problem 13

Solution:

( )

y intercept: x 0

4 0 2y 5

5y

2

=

+ =

=

Problem 14

Solution:

( )

For parallel lines, slopes are equal:

From: 6x 3y 4

6m 2

3

Using Slope-intercept form:

y 2x 6

y 2x 6

+ =

= =

= +

=

Problem 15

Solution:

( ) ( )

m tan tan135

m 1

Using Point-slope form

y 1 1 x 4

y 1 x 4

y x 3

= =

=

=

+ = +

+ =

Problem 16

Solution:

( )y 3 2 x 2

y 3 2x 4

y 2x 1

=

=

=

Problem 17

Solution:

1m

m

4x 2y 1 0

4m 2

2

1 1m

2 2

=

+ =

= =

= =

( ) ( )

Using Point-slope form:

1y 2 x 3

2

2y 4 x 3

x 2y 7 0

=

+ =

=

Problem 18

Solution:The perpendicular bisector passes throughthe midpoint of the line segment.

( )m m

m

m

Midpoint of the line segment: x ,y

2 3 1x

2 2

1 52y

2

+==

= =

( )linesegment

linesegment

Slope :

1 5 6m2 3 5

1 5m

m 6

= =

= =

( )


1 5P , 2 ,m

2 6

5 1y 2 x

6 2

12y 24 10x 5

12y 10x 29

=

=

+ =

=

Problem 19

Solution:Endpoints of the line segment are

the x- and y-intercepts of the line:

x-intercept: y 0=

( )

( )

5x 3 0 15 0

x 3

y-intercept: x 0

5 0 3y 15 0

y 5

+ =

=

=

+ =

=

( ) ( )1 2

linesegment

linesegment

m

m

Endpoints are: P 3,0 &P 0,5

5 0 5m

0 3 3

1 3m

m 5

Perperdicular bisector passes through

the midpoint of the line segment,

solving for the midpoint:

3 0 3x

2 2

0 5 5y

2 2

= =

= =

+= =

+= =


5 3 3y x

2 5 2

3 9 5 9 253y x x

5 10 2 105

3 16y x

5 10

=

+ = + + = +

= +

Problem 20

Solution:

Using Calculator, Equation MODE:

4x 2y 1

x 2y 7

8x

5

+ =

=

=

27y

10

8 27P ,

5 10

=

Problem 21

Solution:

( )

x intyercept: y 0

3x 2 0 12

x 4

=

=

=

Problem 22

Solution:

( ) ( )

4 1 -2 41A

2 5 3 22

120 3 4 2 10 12A

2

A 7.5

=

+ +=

=

Problem 23

Solution:

A(-2,3)

B(1,5)

C(4,2)

A(-2,1)

C(6,-3)

B(4,7)

P(x,y)

A1A2


3/5



& 4th


( ) ( )

( ) ( )

( ) ( )

( ) ( )

total 1 2A A A

6 4 -2 6 4 -2 x 4 6 4 x 61 1 1

-3 7 1 -3 7 1 y 7 -3 7 y -32 2 2

142 4 6 12 14 6

2

14 2y 7x 14 x 4y

2

1 42 4y 3x 12 7x 6y2

36 9 3y 3x 27 y 5x

2x 4y 0 I

Altitud

= +

= +

+ + + =

+ + + +

+ + +

= + +

=

( )

( )

AC

Altitude

AC

Altitude

e BP is perpendicular to AC,

using slope formula:

1 3 4 1m

2 6 8 2

1m 2

m

y 7m 2

x 4

y 7 2x 82x y 1 II

Using calculator:

2x 4y 0 I

2x y 1 II

2x

5

1y

5

2 1P x,y is P ,

5 5

= = =

= =

= =

=

=

=

=

=

=

Problem 24

Solution:

Solution:

( ) ( )

9 7 4 2 91A

3 6 5 2 32

1A 54 35 8 6 21 24 10 18

2

A 15

=

= + + + + + +

=

Problem 25

Solution:

( ) ( ) ( ) ( )

2 21 2

2 2 2 2

2 2 2 2

d d

x 3 y 3 x 4 y 4

x -6x+9+y -6y+9=x -8x+16+y -8y+16

2x 2y 14

x y 7

=

+ = +

+ =

+ =

Problem 26

Solution:y ordinate

x abscissa

y 4x 3

= +

Problem 27

Solution:

( ) ( )( )22

d 0 5 3 6 106= + =

Problem 28

Solution:

( ) ( )

1

2

2 1

1 2

y 3x 2

m 3

y 4x 7

m 4

m mtan

1 m m

4 3tan

1 3 4

4.399

= +

=

= +

=

=

+

=

+

=

Problem 29

Solution:

( )

( ) ( )

( )

1 2

y-intercept: x 0, P 0,-3

P 0, 3 and P 5,2 :

2 3m 1

5 0

=

= =

Problem 30

Solution:

( ) ( )

( )

( )

1P 3,1 and P 10,15

15 1y 1 x 3

10 3

y 1 2 x 3

y 2x 5

=

=

=

Problem 31

Solution:

( )( )

0 2y 0 x 0

0 2

y x

=

=

Problem 32

Solution:

( ) ( ) ( )

( ) ( ) ( )

2 2 2

2 2 2

2 2

x y

2 5 x 3 y 3

2 5 y 3 y 3

20 y 6y 9 y 6y 9

=

= +

= +

= + + + +

22y 2

y 1

=

=

( ) ( )

when y 1:

x 1

when y 1

x 1

Answer : 1, 1 and 1,1

=

=

=

=

Problem 33

Solution:

( )

( )

( ) ( )( ) ( )

( ) ( )

( )

22 2

2 2

2 2

2

x abscissa

y ordinate 4th quadrant

x 3 y

10 x 2 y 4

100 3y 2 y 4

100 9y 12y 4 y 8y 16

10y 20y 80 0

Using calculator:y 4,y 2

Since the point is in the 4th quadrant,

consider, y 2 :

x 3y 3 2

x 6

The p

=

= +

= + +

= + + +

=

= =

=

= =

=

( )oint is P 6,-2 .

Problem 34

Solution:

( )

( )2 2

2

2 2

let point as P x,y :

3 54 2 x y2 2

9 2532 x 3x y 5y

4 4

= +

= + + + + +

( )

2 2

222

2 2

2 2

47x y 3x 5y 0 I

2

9 52 5 x y

2 2

81 2520 x 9x y 5y

4 4

13x y 9x 5y 0 II

2

Subtract II from I:

12x 30 0

5x

2

+ + + =

= +

= + + + +

+ + + =

=

=

2

2

2

5Subs. x in I:

2

5 5 47y 3 5y 0 I

2 2 2

39y 5y 0

4y

=

+ + + =

+ =

A(2,2)B(9,3)

C(7,6)D(4,5)


4/5



& 4th


3 13y ,y

2 2

5 3The point is P ,

2 2

= =

Problem 35

Solution:

( )( ) ( )( )

1 2

1 2 2 1

1 2

r 2, r 3

x r x rx

r r

5 3 6 2x

2 3

3x

5

= =

+=

+

+=

+

=

( ) ( ) ( )( )

1 2 2 1

1 2

y r y ry

r r

4 3 2 2y

2 3

16y

5

3 16The point is P ,5 5

+=

+

+ =

+

=

Problem 36

Solution:

( )( ) ( )( )

( ) ( ) ( )( )

1 2

1 2 2 1

1 2

1 2 2 1

1 2

r 3, r 4

x r x rx

r r

4 4 4 3x

3 4

4x

7

y r y ryr r

5 4 2 3y

3 4

26y

7

4 26The point is ,

7 7

= =

+=

+

+ =

+

=

+=+

+ =

+

=

Problem 37

Solution:

( )( ) ( )( )

1 2 2 1

1 2

2 1

1 2

1 2 2 1

1 2

1

2

1 2

x r x rx

r r

9 r 7 r5

r r

5r 5r 9r 7r

12r 4r

r 1

r 3

r : r 1: 3

+=

+

+ =

+

= +

=

=

=

Problem 38

Solution:

m tan45 1

9 5y 1 x

2 2

9 5y x

2 2

Solving for x when y 6 :

9 5y x

2 2

9 56 x

2 2

x 8 abscissa

= =

=

+ = +

=

+ = +

+ = +

=

Problem 39

Solution:

2 1

1 2

11

1 1

2

2

2 2

2

m mtan

1 m m

1tan

2

1tan m

2

1m 2tan45

11 m

2

1 11 m m

2 2

m 3

=

+

=

= =

=

+

+ =

=

Problem 40

Solution:

( )

( )

( )

1

2

1 2

1

2

7 7 14m

x 6 x 6

2 4

m 15 1

m m

1m

m

14 1

x 6 1

14 x 6

x 8

= =

+

= =

=

= +

= +

=

Problem 41

Solution:

( )

( )

( )

( )

Perpendicular bisector passes through themidpoint and is perpendicular to the line

segment. Solving for the midpoint:

4 6x 1

2

0 3 3y

2 2

0 3 3m

4 6 10

10m

3

+ = =

+ = =

= =

=

( )( )

Perpendicular bisector:

3 10P 1, ,m

2 3

3 10y x 1

2 3

6y 9 20x 20

20x 6y 29 0

=

=

+ =

+ + =

Problem 42

Solution:

( )

arctan2

m tan 2

y 9 2 x 6

y 9 2x 12

2x y 3 0

=

= =

=

=

=

Problem 43

Solution:

( ) ( )( )

( )

( )

( )( )

( )

( )

1

2

d equation:

2 1y 2 x 88 4

1y 2 x 8

12

12y 24 x 8

x 12y 16 I

d equation:

1 7y 7 x 0

2 0

y 7 4 x 0

4x y 7 II

Point of intersection, I in II,

using calculator

68x

49

71y

49

68 71P ,

49 49

=

+ =

+ = +

+ =

=

+ =

=

=

=

Problem 44

Solution:

( )

( )

1

2

2 1

1 2

2x y 8 0

m 2

x 3y 4 01

m3

m mtan

1 m m

12

3tan1

1 23

tan 1

45

+ =

=

+ + =

=

=

+

=

+

=

=


5/5



& 4th


Problem 45

Solution:

( )

( )

8arctan

5

8m tan

5

P 0,0 origin

8y 0 x 0

5

5y 8x

8x 5y 0

=

= =

=

=

+ =

Problem 46

Solution:

A line touching the circle is the line tangent tothe circle and is perpendicular to the locus ofthe center of a circle.

( )

( )

8x y 3 0

m 81

m8

P 1,5

1y 5 x 1

8

8y 40 x 1

x 8y 41 0

=

=

=

=

= +

+ =

Problem 47

Solution:Points on the line:

( ) ( ) ( )( )

( )

( )

( )( )

2

2

7,4 , x,0 & 0, x0 x4 0

m7 x x 0

4x 7x x

x 11x

x 11

4 0 4m 1

7 11 4

P 7,4 ,m 1

y 4 1 x 7

y 4 x 7

x y 11 0

= =

=

=

=

= = =

=

=

= +

+ =

Problem 48

Solution:

( ) ( )3 8

, , x,0 & 0,x5 5

80

0 x5m3 x 0x5

= =

28 3x x x5 5

x 1

80

5m 13

15

= +

=

= =

3 8P , ,m 1

5 538

y 1 x55

8 3y x

5 5

5y 8 5x 3

5x 5y 5 0

x y 1 0

=

=

=

=

+ =

+ =

Problem 49

Solution:

( )

( )

( )

( ) ( )

( )( )( )

( )

22 2

2

2 2

x,0 x intercept

0,y y intercept

x y 7 2

y 98 x I

5 y 5 0m

2 0 2 x

10 2y 5x xy 10

2y 5x xy 0 II

I in II:

2 98 x 5x x 98 x 0

Using calculator:

x 7

Using two point form:

-2,5 & 7,0

5 0y 5 x 2

2 7

y 5 x 2

x y 7

+ =

=

= =

+ + =

+ =

+ =

=

=

= +

+ = 0

Problem 50

Solution:

( )

( )

( ) ( )

2

2

yx,0 ,0 x intercept

2

0,y y interceptPoints are:

y-2,6 , ,0 & 0,y

2

6 0 y 0m

y y2 0

2 2

6y y2y

2 2

6y 4y y

y 2

=

= =

=

=

=

( ) ( )

( )( )

Using two-point form:

-2,6 & 0,2

6 2y 6 x 2

2 0

y 6 2x 4

2x y 2 0

=

=

+ =

analytic 1.pdf

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