# Free Objective Test 02 Practice Test - 11th and 12th

If the vertices of a triangle be (a, 1), (b, 3) and (4, c), then the centroid of the triangle will lie on x-axis, if

A.

a + c = -4

B.

a + b = -4

C.

c = -4

D.

b + c = -4

#### SOLUTION

Solution : C

The point lies on axis of x, if y = 0.

Therefore,  1+3+c3 = 0 c = -4.

If the vertices of a triangle be (a, b - c), (b, c - a) and (c, a - b), then the centroid of the triangle lies

A.

At origin

B.

On x-axis

C.

On y-axis

D.

(a+b+c,0)

#### SOLUTION

Solution : B

x =  a+b+c3, y =  bc+ca+ab3 = 0

Hence, centroid lies on x - axis.

Without changing the direction of coordinate axes, origin is transferred to (h, k), so that the linear (one degree)

terms in the equation x2+y24x+6y7=0 are eliminated. Then the point (h, k) is

A.

(3, 2)

B.

(- 3, 2)

C.

(2, - 3)

D.

(1.7)

#### SOLUTION

Solution : C

Putting x = x' + h, y = y' + k, the given equation transforms to

x2+y2+x(2h4)+y(2k+6)+h2+k27=0

To eliminate linear terms, we should have

2h - 4 = 0, 2k + 6 = 0  h = 2, k = -3

i.e., (h, k) = (2, -3).

The orthocenter of the triangle formed by (0, 0), (8, 0) and (4, 6) is

A. 4,83
B. 3,4
C. 4,3
D. 3,4

#### SOLUTION

Solution : A

Sol: Let A ≡ (0, 0), B ≡ (8, 0) and C ≡ (4, 6).

Slope of BC = 6040=32

Equation of the line through A(0, 0) and perpendicular to BC is
(y – 0) = 23 (x – 0) i.e. 2x – 3y = 0                                                             …… (1)

Slope of CA =6040=32

Equation of the line through B(8, 0) and perpendicular to CA is

(y – 0) = 23 (x – 8) i.e., 2x + 3y = 16                                                         …… (2)

Solving (1) and (2), the orthocenter is  4,83

The equations of two equal sides of an isosceles triangle are 7x – y + 3 = 0 and x + y –3 = 0 and the third side passes through the point (1, -10). The equation of the third side is

A. x – 3y – 31 = 0 but not 3x + y + 7 = 0
B. 3x + y + 7 = 0 but not x – 3y – 31 = 0
C. 3x + y + 7 = 0 or  x – 3y – 31 = 0
D. Neither 3x + y + 7 nor x – 3y – 31 = 0

#### SOLUTION

Solution : C

Any line through (1, - 10) is given by y + 10 = m(x - 1)
Since it makes equal angle α with the given lines 7x – y + 3 = 0 and x + y – 3 = 0, therefore
tanα=m71+7m=m+11+m(1)m=13 or 3
Hence the two possible equations of third side are 3x + y + 7 = 0 and x - 3y - 31 = 0.

Let PS be the median of the triangle with vertices P(2, 2), Q(6, -1) and R(7, 3). The equation of the line passing through (1, -1) and parallel to PS is

A. 2x – 9y – 7 = 0
B. 2x – 9y – 11 = 0
C. 2x + 9y – 7 = 0
D. 2x – 9y + 7 = 0

#### SOLUTION

Solution : D

S = midpoint of QR = (6+72,1+32)=(132,1)slope of PS=212132=29The required equation is y+1=29(x1)i.e.,2x+9y+7=0

If the co-ordinates of the middle point of the portion of a line intercepted between coordinate axes is (3, 2), then the equation of the line will be

A. 2x + 3y = 12
B. 3x+2y=12
C. 4x - 3y = 6
D. 5x - 2y = 10

#### SOLUTION

Solution : A

Using midpoint formula we get the coordinates of the points A and B as (6, 0) and (0, 4) respectively.

Therefore, using intercept form of the striaght line,  the equation of line AB is x6+y4=1
2x+3y=12

Let A (h, k), B (1, 1) and C (2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which `k' can take is given by

A. {1, 3}
B. {0, 2}
C. {-1, 3}
D. {-3, -2}

#### SOLUTION

Solution : C

Since, A(h, k), B(1, 1) and C (2, 1) are the vertices of a right angled ΔABC.

Now, area of ΔABC
=12|k1|.1
1=12|k1|
k1=±2
k=1,3

If (α, β), (¯x  , ¯y) and (u, v)are respectively coordinates of the circumcentre, centroid and orthocentre of a triangle.

A. 3¯x =2α+u and 3¯y =2β+v
B. 3¯x =2αu and 3¯y =2βv
C. 3¯x =2αu and 3¯y =2β+v
D. 3¯x =2α+u and 3¯y =2βv

#### SOLUTION

Solution : C

We know that, the centroid of a triangle divides the segment joining the orthocentre and circumcentre internally in the ratio 2 : 1. Therefore,
¯x=2α+u2+1and¯y=2β+v2+1
3¯x=2α+uand¯y=2β+v

The distance between the parallel lines 8x+6y+5=0 and 4x+3y25=0 is

A.

73

B.

92

C.

112

D.

54

#### SOLUTION

Solution : C

Distance between parallel lines ax+by+c1=0 and ax+by+c2=0 is |c1c2|a2+b2

Given lines are 8x+6y+5=0 and
4x+3y25=0 or 8x+6y50=0

Required distance = 5+5082+62=5510=112

The angle between the pair of straight lines x2+4y27xy=0, is

A.

tan113

B.

tan13

C.

tan1335

D.

tan1533

#### SOLUTION

Solution : C

tanθ=2h2aba+b
θ=tan1249445=tan1335.

If (a + 3b)(3a + b) = 4h2, then the angle between the lines represented by ax2+2hxy+by2=0 is

A.

90

B.

45

C.

60

D.

tan112

#### SOLUTION

Solution : C

θ=tan1(2h2aba+b)=tan1(4h24aba+b)

= tan1(3a2+3b2+10ab4aba+b)=60.

O is the origin and A is the point (3,4). If a point P moves so that the line segment OP is always parallel to the line segment OA, then the equation to the locus of P is

A.

4x - 3y = 0

B.

4x + 3y = 0

C.

3x + 4y = 0

D.

3x - 4y = 0

#### SOLUTION

Solution : A

Since OA and OP will be parallel only when O, A and P are collinear.

Therefore,  ∣ ∣001341xy1∣ ∣ = 0 4x - 3y = 0.

A point P moves so that its distance from the point (a, 0) is always equal to its distance from the line x + a = 0. The

locus of the point is

A.

y2=4ax

B.

x2=4ay

C.

y2+4ax=0

D.

x2+4ay=0

#### SOLUTION

Solution : A

(xa)2+y2=(x+a)2y2=4ax

Note: This is also the definition of parabola y2 = 4ax.

If P = (1, 0), Q = (-1, 0) and R = (2, 0) are three given points, then the locus of a point S satisfying the relation

SQ2+SR2=2SP2 is

A.

A straight line parallel to x-axis

B.

A circle through origin

C.

A circle with centre at the origin

D.

A straight line parallel to y-axis

#### SOLUTION

Solution : D

Let S(x, y), then

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

2x +1 + 4 - 4x = - 4x + 2 x = -32

Hence it is a straight line parallel  to y-axis.