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

### Question 1

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 + c = -4

a + b = -4

c = -4

b + c = -4

#### SOLUTION

Solution :C

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

Therefore, 1+3+c3 = 0 ⇒ c = -4.

### Question 2

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

At origin

On *x*-axis

On *y*-axis

(a+b+c,0)

#### SOLUTION

Solution :B

x = a+b+c3, y = b−c+c−a+a−b3 = 0

Hence, centroid lies on x - axis.

### Question 3

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

terms in the equation x2+y2−4x+6y−7=0 are eliminated. Then the point (h, k) is

(3, 2)

(- 3, 2)

(2, - 3)

(1.7)

#### SOLUTION

Solution :C

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

x′2+y′2+x′(2h−4)+y′(2k+6)+h2+k2−7=0

To eliminate linear terms, we should have

2h - 4 = 0, 2k + 6 = 0 ⇒ h = 2, k = -3

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

### Question 4

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

#### SOLUTION

Solution :A

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

Slope of BC = 6−04−0=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 =6−04−0=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

### Question 5

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

#### 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=m−71+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.

### Question 6

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

#### SOLUTION

Solution :D

S = midpoint of QR = (6+72,−1+32)=(132,1)∴slope of PS=2−12−132=−29∴The required equation is y+1=−29(x−1)i.e.,2x+9y+7=0

### Question 7

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

#### 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

### Question 8

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

#### 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|k−1|.1

⇒1=12|k−1|

⇒k−1=±2

⇒k=−1,3

### Question 9

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

#### 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

### Question 10

The distance between the parallel lines 8x+6y+5=0 and 4x+3y−25=0 is

73

92

112

54

#### SOLUTION

Solution :C

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

Given lines are 8x+6y+5=0 and

4x+3y−25=0 or 8x+6y−50=0∴ Required distance = 5+50√82+62=5510=112

### Question 11

The angle between the pair of straight lines x2+4y2−7xy=0, is

tan−113

tan−13

tan−1√335

tan−15√33

#### SOLUTION

Solution :C

tanθ=2√h2−aba+b

θ=tan−12√494−45=tan−1√335.

### Question 12

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

90∘

45∘

60∘

tan−112

#### SOLUTION

Solution :C

θ=tan−1(2√h2−aba+b)=tan−1(√4h2−4aba+b)= tan−1(√3a2+3b2+10ab−4aba+b)=60∘.

### Question 13

*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

4x - 3y = 0

4x + 3y = 0

3x + 4y = 0

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.

### Question 14

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

y2=4ax

x2=4ay

y2+4ax=0

x2+4ay=0

#### SOLUTION

Solution :A

(x−a)2+y2=(x+a)2⇒y2=4ax

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

### Question 15

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 straight line parallel to *x*-axis

A circle through origin

A circle with centre at the origin

A straight line parallel to *y*-axis

#### SOLUTION

Solution :D

Let S(x, y), then

(x+1)2+y2+(x−2)2+y2=2[(x−1)2+y2]

⇒ 2x +1 + 4 - 4x = - 4x + 2 ⇒ x = -32

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