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

### Question 1

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

(-3,3)

(3,3)

(3,1)

(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=3y=−3−2+83=1

### Question 2

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

(4,8)

(8,4)

(6,4)

(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=6Centroid (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).

### Question 3

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

#### SOLUTION

Solution :C

Let AD be the internal bisector of angle BAC cutting BC at D.Now, AB=√(5−2)2+(2−3)2=√10and AC=√(5−6)2+(2−5)2=√10

since AD is the internal bisector of angle BAC,

∴ BDDC=ABAC=√10√10=11

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

So, the equation of AD is

y−2=2−45−4 (x – 5) or 2x + y – 12 = 0

### Question 4

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

(2−√3)x−y−4+2√3=0

(2−√3)x+y−4+2√3=0

(2−√3)x−y+4+2√3=0

(2−√3)x−9y+4+2√3=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∘=(2−√3)

∴ Equation of the line AC is

(y−0)=(2−√3)(x−2)or y=(2−√3)x−4+2√3or (2−√3)x−y−4+2√3=0

### Question 5

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

27

15

18

7

#### SOLUTION

Solution :D

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

= 12 [56 - 60 + 18] = 7.

### Question 6

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

(2,1)

(−1,4)

(1,2)

(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

x−x1a=y−y1b=−(ax1+by1+c)a2+b2

Here (x1,y1)=(2,3);ax+by+c=x+y−3

∴ x−21=y−31=−((1×2)+(1×3)+(−3))12+12

⇒ x−21=y−31=−((2)+(3)+(−3))2

⇒ x−21=y−31=−22

⇒ x−21=y−31=−1

∴ x - 2 = -1 ; y-3 = -1

⇒ x = -1 + 2 ; y = -1 + 3

⇒ x = 1 ; y = 2

∴ (x,y) = (1,2)

### Question 7

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

(23,56)

(13,12)

(12,43)

(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)

### Question 8

If the line

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

is parallel to the y-axis, then the value of k is

#### 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

### Question 9

The angle between the pair of straight lines x2−y2−2y−1=0, is

90∘

60∘

75∘

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

### Question 10

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

p = q

q = 0

p = 0

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.

### Question 11

The lines joining the points of intersection of line x + y = 1 and curve x2+y2−2y+λ=0 to the origin are perpendicular, then the value of λ will be

12

−12

1√2

0

#### SOLUTION

Solution :D

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

x2+y2−2y(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+λ−1+λ=0⇒λ=0

### Question 12

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

12x2+4y2=3

12x2−4y2=3

12x2−4y2+3 = 0

12x2+4y2+3 = 0

#### SOLUTION

Solution :B

According to the given condition

√(x−1)2+y2−√(x+1)2+y2 = ±1

On squaring both sides, we get

2x2+2y2+1=2√(x−1)2+y2√(x+1)2+y2

Again on squaring, we get 12x2−4y2=3.

### Question 13

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

k−4h+3× 4=−1

⇒ 4k - 16 = -h - 3

⇒ h + 4k = 13....(i)

and slope of PB× slope of AC = - 1

⇒k−85h+35×−23=−1

⇒5k−85h+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.

### Question 14

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

#### 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=k2⇒y2=2x

### Question 15

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

#### 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).