# Free Objective Test 01 Practice Test - 11th and 12th

If A(4, -3), B(3, -2) and C(2, 8) are the vertices of a triangle, then its centroid will be

A.

(-3,3)

B.

(3,3)

C.

(3,1)

D.

(1,3)

#### SOLUTION

Solution : C

Let the centroid of the triangle be (x, y).

The centroid of a triangle is given by (x1+x2+x33,y1+y2+y33)

x=4+3+23=3

y=32+83=1

If two vertices of a triangle are (6,4), (2,6) and its centroid is (4, 6), then the third vertex is

A.

(4,8)

B.

(8,4)

C.

(6,4)

D.

(0,0)

#### SOLUTION

Solution : A

Given:
Centroid = (4,6)
Vertices (6,4) & (2,6)

Let the Co-ordinates of C be (x3, y3)
x1=6, x2=32, y1=4 & y2=6

Centroid (4,6)=(x1+x2+x33,y1+y2+y33)

4=6+2+x33 and  6=4+6+y33
x3=4 and  y3=8

Third vertex is (4,8).

The equation of the internal bisector of BAC of ΔABC with vertices A(5, 2), B(2, 3) and C(6, 5), is

A. 2x + y + 12 = 0
B. x + 2y – 12 = 0
C. 2x + y – 12 = 0
D. x + 2y +12 = 0

#### SOLUTION

Solution : C

Let AD be the internal bisector of angle BAC cutting BC at D.

Now,  AB=(52)2+(23)2=10and  AC=(56)2+(25)2=10

since AD is the internal bisector of angle BAC,

BDDC=ABAC=1010=11

Coordinates of D are (2+62,3+52) i.e. (4, 4)

So, the equation of AD is

y2=2454 (x – 5) or 2x + y – 12 = 0

A line through the point A(2, 0), which makes an angle of 30 with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15. The equation of the straight line in the new position is

A.

(23)xy4+23=0

B.

(23)x+y4+23=0

C.

(23)xy+4+23=0

D.

(23)x9y+4+23=0

#### SOLUTION

Solution : A

Let AB be the initial position of the line and AC be its new position.

Slope of the line

AC=tan 15=(23)

Equation of the line AC is

(y0)=(23)(x2)or y=(23)x4+23or (23)xy4+23=0

If the coordinates of the points A, B, C, be (4,4), (3,-2) and (3,-16) respectively, then the area of the triangle ABC is

A.

27

B.

15

C.

18

D.

7

#### SOLUTION

Solution : D

= 12[4(- 2 + 16) + 3(-16 - 4) + 3(4 + 2)]

=  12 [56 - 60 + 18] = 7.

If PM is the perpendicular from P (2,3) on to the line x+y=3 then the co-ordinates of M are

A.

(2,1)

B.

(1,4)

C.

(1,2)

D.

(4,1)

#### SOLUTION

Solution : C

If (x,y) is the foot of the perpendicular from (x1,y1) to the line ax + by + c = 0 then

xx1a=yy1b=(ax1+by1+c)a2+b2

Here (x1,y1)=(2,3);ax+by+c=x+y3

x21=y31=((1×2)+(1×3)+(3))12+12

x21=y31=((2)+(3)+(3))2

x21=y31=22

x21=y31=1

x - 2 = -1 ; y-3 = -1

x = -1 + 2 ; y = -1 + 3

x = 1 ; y = 2

(x,y) = (1,2)

If 4a+5b+6c=0 then the set of lines ax+by+c=0 are concurrent at the point

A.

(23,56)

B.

(13,12)

C.

(12,43)

D.

(13,73)

#### SOLUTION

Solution : A

4a+5b+6c=0
(6x)a+(6y)b+6c=0  [Multiply given set of lines ax+by+c=0 with '6']
Now on comparing 6x=4 and 6y =5
(x,y) =  (23,56)
ax+by+c=0 must passes through (23,56)

If the line
(3x+14y+7)+k(5x+7y+6)=0
is parallel to the y-axis, then the value of k is

A. 13
B. 35
C. 2
D. 2

#### SOLUTION

Solution : C

Given line is (3x+14y+7)+k(5x+7y+6)=0

(3+5k)x+(14+7k)y+(7+6k)=0

If it is parallel to y- axis, then coefficient of y =0

14+7k=0

k=2

The angle between the pair of straight lines x2y22y1=0, is

A.

90

B.

60

C.

75

D.

36

#### SOLUTION

Solution : A

Pair of straight lines represented by a second degree equation with coefficient of x2  as a and coefficient of y2  as b are perpendicular if a+b = 0
Here a + b = 0, so they are perpendicular to each other

If the lines (pq)x2+2(p+q)xy+(qp)y2=0 are mutually perpendicular, then

A.

p = q

B.

q = 0

C.

p = 0

D.

p and q may have any value

#### SOLUTION

Solution : D

Pair of straight lines represented by a second degree equation with coefficient of x2  as a and coefficient of y2  as b are perpendicular if a+b = 0. Herere a + b = 0 for every p and q.

The lines joining the points of intersection of line x + y = 1 and curve x2+y22y+λ=0 to the origin are perpendicular, then the value of λ will be

A.

12

B.

12

C.

12

D.

0

#### SOLUTION

Solution : D

Making the equation of curve homogeneous with the help of line x + y =1,we get

x2+y22y(x+y)+λ(x+y)2=0

x2(1+λ)+y2(1+λ)2yx=0

Therefore the lines be perpendicular, if A + B = 0.

1+λ1+λ=0λ=0

Two points A and B have coordinates (1, 0) and (-1, 0) respectively and Q is a point which satisfies the relation

AQ - BQ = ± 1. The locus of Q is

A.

12x2+4y2=3

B.

12x24y2=3

C.

12x24y2+3 = 0

D.

12x2+4y2+3 = 0

#### SOLUTION

Solution : B

According to the given condition

(x1)2+y2(x+1)2+y2 = ±1

On squaring both sides, we get

2x2+2y2+1=2(x1)2+y2(x+1)2+y2

Again on squaring, we get 12x24y2=3.

The orthocenter of the triangle formed by the lines x + y = 1, 2x + 3y = 6 and 4x - y + 4 = 0 lies in

#### SOLUTION

Solution : A

Coordinates of A and B are (-3, 4) and (35,85)​ if orthocenter P(h, k)

Then, (slope of PA)× (slope of BC) = - 1
k4h+3× 4=1

4k - 16 = -h - 3
h + 4k = 13....(i)

and slope of PB× slope of AC = - 1
k85h+35×23=1
5k85h+3×23=1
10k - 16 = 15th + 9
15th - 10k + 25 = 10
3h - 2k + 5 = 0     ...(ii)
Solving Eqs. (i) and (ii), we get h=37,k=227
Hence, orthocentre lies in I quadrant.

If h denote the A.M, k denote G.M of the intercepts made on axes by the lines passing through (1, 1) then (h, k) lies on

A. y2=2x
B. y2=4x
C. y=2x
D. x+y=2xy

#### SOLUTION

Solution : A

a = x - intercept, b = y - intercept
2h=a+b, k2=ab
xa+yb=1​, substitute (1, 1)
1a+1b=1
a + b = ab
2h=k2y2=2x

If the orthocenter and circumcentre of a triangle are (0,0) and (3,6) respectively then the centroid of the triangle is

A. (1,2)
B. (2,4)
C. (23,43)
D. (13,23)

#### SOLUTION

Solution : B

In any triangle centroid divides the line joining orthocenter and circumcentre internally in the ratio 2 : 1.
Applying section formula to find the point which divides the line joining (0,0) in the ratio 2:1 , we get the coordinated of centroid equal to (2,4).