-7) 2) (3,-2) EXERCISE-2 (LOCUS) LEVEL-I us of the point, the sum of whose distances from the coordinate axes is 9 is 3) lyl- x = 9 1?) and (3,0) is 4) y+ x = 9 2) Ixl + lyl = 9 4) x + v-3.

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Question
quilateral triangle
ght angled triangle
o vertices o
0,10) and B = (30,20) be two points and P (x,0) be a point on the x-axis. The value
4) 10
3) 4
integral coordinates is
2)5
The
13.
1) x+2y
4) 3:4
3) 2:1
perpend
1)x
3) x+
longer side is
14 A(2.3) a
2) 4:3
a-B
15. If A=
1) x
a-p
3) cos
4) sin
a-8
2) cos
2.
V3
2
16. Let A
9x + 7
1) AB
2.
3) (-2,3)
4) (5,–1)
-7)
2) (3,-2)
17. The
The
EXERCISE-2 (LOCUS)
LEVEL-I
cus of the point, the sum of whose distances from the Coordinate axes is
1) 6
18. The
is
lyl 9
4) y + x = 9
2) |x| + ly] = 9
cus of the point equidistant from (-1,2) and (3,0) is
2) 2x-y + 1 = 0
%3D
1)
y-1 0
3) x + y +1 = 0
4) x + y – 2 = 0
19. Le
%3D
cus of the point which is at a constant distance of 5 from the point (2.3) is
1.
y-4x-6y-25 = 0
y²-4x-6y-8 = 0
2) x + y² – 4x - 6y + 25 = 0
4) x² + y² – 4x – 6y – 12 = 0
20. If
is
-y 0
us of the point P from which the distance to (4,0) is double the distance from P to
is
By2-8x+16 0
y - 8y + 16 = 0
us of the point P which moves such that its distance from (4,0) is twice its distance
line x = 1 is
By 12
9,0) and B (-1,0) then the locus of P such that PA = 3PB is
2) x- 3y² = 0
%3D
3) 3x? - y = 0
%3D
4) x²– 2y² = 0
%3D
%3D
21.
%3D
2) x²- 3y² - 8x- 16 = 0
4) 3x -y-8y- 16 = 0
22.
%3D
%3D
%3D
2) 3x- y = 12
23
%3D
3) x + 3y? = 12
2) y -x = 9
4) 3x² + y = 12
%3D
%3D
3) x² + y² = 9
%3D
which the sum of the distance of P from A andB is minimum equals
2
%3D
4) x² - y²+9 = 0
2)20
a
enal Institutes
3) 15
4)0
1n
Transcribed Image Text:quilateral triangle ght angled triangle o vertices o 0,10) and B = (30,20) be two points and P (x,0) be a point on the x-axis. The value 4) 10 3) 4 integral coordinates is 2)5 The 13. 1) x+2y 4) 3:4 3) 2:1 perpend 1)x 3) x+ longer side is 14 A(2.3) a 2) 4:3 a-B 15. If A= 1) x a-p 3) cos 4) sin a-8 2) cos 2. V3 2 16. Let A 9x + 7 1) AB 2. 3) (-2,3) 4) (5,–1) -7) 2) (3,-2) 17. The The EXERCISE-2 (LOCUS) LEVEL-I cus of the point, the sum of whose distances from the Coordinate axes is 1) 6 18. The is lyl 9 4) y + x = 9 2) |x| + ly] = 9 cus of the point equidistant from (-1,2) and (3,0) is 2) 2x-y + 1 = 0 %3D 1) y-1 0 3) x + y +1 = 0 4) x + y – 2 = 0 19. Le %3D cus of the point which is at a constant distance of 5 from the point (2.3) is 1. y-4x-6y-25 = 0 y²-4x-6y-8 = 0 2) x + y² – 4x - 6y + 25 = 0 4) x² + y² – 4x – 6y – 12 = 0 20. If is -y 0 us of the point P from which the distance to (4,0) is double the distance from P to is By2-8x+16 0 y - 8y + 16 = 0 us of the point P which moves such that its distance from (4,0) is twice its distance line x = 1 is By 12 9,0) and B (-1,0) then the locus of P such that PA = 3PB is 2) x- 3y² = 0 %3D 3) 3x? - y = 0 %3D 4) x²– 2y² = 0 %3D %3D 21. %3D 2) x²- 3y² - 8x- 16 = 0 4) 3x -y-8y- 16 = 0 22. %3D %3D %3D 2) 3x- y = 12 23 %3D 3) x + 3y? = 12 2) y -x = 9 4) 3x² + y = 12 %3D %3D 3) x² + y² = 9 %3D which the sum of the distance of P from A andB is minimum equals 2 %3D 4) x² - y²+9 = 0 2)20 a enal Institutes 3) 15 4)0 1n
Let A (0,10) and B (30,20) be two points and P (x,0) be a point on the x-axis. The value
110. if the vertices of a triangle
1) equilateral triangle
3) right angled triangle
Let A (3, 4) and B is variable point on the lines px
with integral coordinates is
1) 6
4) 10
3) 4
2)5
13.
4) 3:4
3) 2:1
14
to the longer side is
1) 1:2
2) 4:3
1
if
4) sin -B
a-B.
3) cos
15.
a-B
2) cos
V3
2.
1) sin
%3D
16.
%3D
is
3) (-2,3)
4) (5,–1)
1) (-4,-7)
2) (3,-2)
17
EXERCISE-2 (LOCUS)
LEVEL-I
14
3) Iyl-지 = 9
4) y + x = 9
%3D
1) x- lyl = 9
The locus of the point equidistant from (-1,2) and (3,0) is
1) 2x-y-1= 0
The locus of the point which is at a constant distance of 5 from the point (2,3) is
1) x² + y - 4x- 6y-25 = 0
3) x² + y? -4x- 6y - 8 = 0
A point moves so that its distance from y-axis is half of its distance from the origin. The
of point is
1) 2x – y = 0
The locus of the point P from which the distance to (4,0) is double the distance from P to a
X- axis is
1) x – 3y - 8x + 16 = 0
3) 3x - y - 8y + 16 = 0
The locus of the point P which moves such that its distance from (4,0) is twice its distance
from the line x = 1 is
2) 지 + lyl = 9
2) 2x - y+ 1 = 0
3) x + y + 1 = 0
4) x + y-2 = 0
2) x² + y? – 4x - 6y + 25 = 0
4) x² + y² – 4x- 6y - 12 = 0
%3D
2) x²- 3y 0
3) 3x - y = 0
%3D
%3D
4) x²– 2y = 0
%3D
%3D
2) x²- 3y - 8x-16 0
4) 3x-y-8y- 16 = 0
%3D
%3D
1) x-3y = 12
fA=(-9,0) and B = (-1,0) then the locus of P such that PA = 3PB is
1) x-y 9
2) 3x -y 12
%3D
%3D
3) x² + 3y = 12
%3D
4) 3x² + y = 12
%D
2) y-x 9
%3D
%3D
3) x² + y² = 9
f x for which the sum of the distance of P from A and B is minimum equals
) 10
%3D
%3D
4) x² -y+ 9= 0
2) 20
lucational Institutes
3)15
4) 0
Transcribed Image Text:Let A (0,10) and B (30,20) be two points and P (x,0) be a point on the x-axis. The value 110. if the vertices of a triangle 1) equilateral triangle 3) right angled triangle Let A (3, 4) and B is variable point on the lines px with integral coordinates is 1) 6 4) 10 3) 4 2)5 13. 4) 3:4 3) 2:1 14 to the longer side is 1) 1:2 2) 4:3 1 if 4) sin -B a-B. 3) cos 15. a-B 2) cos V3 2. 1) sin %3D 16. %3D is 3) (-2,3) 4) (5,–1) 1) (-4,-7) 2) (3,-2) 17 EXERCISE-2 (LOCUS) LEVEL-I 14 3) Iyl-지 = 9 4) y + x = 9 %3D 1) x- lyl = 9 The locus of the point equidistant from (-1,2) and (3,0) is 1) 2x-y-1= 0 The locus of the point which is at a constant distance of 5 from the point (2,3) is 1) x² + y - 4x- 6y-25 = 0 3) x² + y? -4x- 6y - 8 = 0 A point moves so that its distance from y-axis is half of its distance from the origin. The of point is 1) 2x – y = 0 The locus of the point P from which the distance to (4,0) is double the distance from P to a X- axis is 1) x – 3y - 8x + 16 = 0 3) 3x - y - 8y + 16 = 0 The locus of the point P which moves such that its distance from (4,0) is twice its distance from the line x = 1 is 2) 지 + lyl = 9 2) 2x - y+ 1 = 0 3) x + y + 1 = 0 4) x + y-2 = 0 2) x² + y? – 4x - 6y + 25 = 0 4) x² + y² – 4x- 6y - 12 = 0 %3D 2) x²- 3y 0 3) 3x - y = 0 %3D %3D 4) x²– 2y = 0 %3D %3D 2) x²- 3y - 8x-16 0 4) 3x-y-8y- 16 = 0 %3D %3D 1) x-3y = 12 fA=(-9,0) and B = (-1,0) then the locus of P such that PA = 3PB is 1) x-y 9 2) 3x -y 12 %3D %3D 3) x² + 3y = 12 %3D 4) 3x² + y = 12 %D 2) y-x 9 %3D %3D 3) x² + y² = 9 f x for which the sum of the distance of P from A and B is minimum equals ) 10 %3D %3D 4) x² -y+ 9= 0 2) 20 lucational Institutes 3)15 4) 0
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