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 $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

$x = \frac{4 + 3 + 2}{3} = 3$

$y = \frac{- 3 - 2 + 8}{3} = 1$

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 $(x_{3}, y_{3})$
$x_{1} = 6, x_{2} = 32, y_{1} = 4 & y_{2} = 6$

Centroid $(4, 6) = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

$\Rightarrow 4 = \frac{6 + 2 + x_{3}}{3}$ and $6 = \frac{4 + 6 + y_{3}}{3}$
$\Rightarrow$ $x_{3} = 4$ and $y_{3} = 8$

$∴$ Third vertex is $(4, 8)$ .

Question 3

The equation of the internal bisector of $∠$ BAC of $Δ$ ABC 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.
$N o w, A B = \sqrt{(5 - 2)^{2} + (2 - 3)^{2}} = \sqrt{10} a n d A C = \sqrt{(5 - 6)^{2} + (2 - 5)^{2}} = \sqrt{10}$

since AD is the internal bisector of angle BAC,

$∴ \frac{B D}{D C} = \frac{A B}{A C} = \frac{\sqrt{10}}{\sqrt{10}} = \frac{1}{1}$

$∴$ Coordinates of D are $(\frac{2 + 6}{2}, \frac{3 + 5}{2})$ i.e. (4, 4)

So, the equation of AD is

$y - 2 = \frac{2 - 4}{5 - 4}$ (x – 5) or 2x + y – 12 = 0

Question 4

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

$(2 - \sqrt{3}) x - y - 4 + 2 \sqrt{3} = 0$

$(2 - \sqrt{3}) x + y - 4 + 2 \sqrt{3} = 0$

$(2 - \sqrt{3}) x - y + 4 + 2 \sqrt{3} = 0$

$(2 - \sqrt{3}) x - 9 y + 4 + 2 \sqrt{3} = 0$

SOLUTION

Solution : A

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

Slope of the line

$A C = t a n 15^{\circ} = (2 - \sqrt{3})$

$∴$ Equation of the line AC is

$(y - 0) = (2 - \sqrt{3}) (x - 2) o r y = (2 - \sqrt{3}) x - 4 + 2 \sqrt{3} o r (2 - \sqrt{3}) x - y - 4 + 2 \sqrt{3} = 0$

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

SOLUTION

Solution : D

$△$ = $\frac{1}{2}$ [4(- 2 + 16) + 3(-16 - 4) + 3(4 + 2)]

= $\frac{1}{2}$ [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 $(x_{1}, y_{1})$ to the line ax + by + c = 0 then

$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{- (a x_{1} + b y_{1} + c)}{a^{2} + b^{2}}$

Here $(x_{1}, y_{1}) = (2, 3); a x + b y + c = x + y - 3$

$∴$ $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{- ((1 \times 2) + (1 \times 3) + (- 3))}{1^{2} + 1^{2}}$

$\Rightarrow$ $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{- ((2) + (3) + (- 3))}{2}$

$\Rightarrow$ $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{- 2}{2}$

$\Rightarrow$ $\frac{x - 2}{1} = \frac{y - 3}{1} = - 1$

$∴$ x - 2 = -1 ; y-3 = -1

$\Rightarrow$ x = -1 + 2 ; y = -1 + 3

$\Rightarrow$ 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

$(\frac{2}{3}, \frac{5}{6})$

$(\frac{1}{3}, \frac{1}{2})$

$(\frac{1}{2}, \frac{4}{3})$

$(\frac{1}{3}, \frac{7}{3})$

SOLUTION

Solution : A

4a+5b+6c=0
(6x)a+(6y)b+6c=0 $\to$ [Multiply given set of lines ax+by+c=0 with '6']
Now on comparing 6x=4 and 6y =5
⇒ (x,y) = $(\frac{2}{3}, \frac{5}{6})$
$∴$ ax+by+c=0 must passes through $(\frac{2}{3}, \frac{5}{6})$

Question 8

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

\frac{1}{3}

\frac{- 3}{5}

- 2

2

SOLUTION

Solution : C

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

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

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

$\Rightarrow 14 + 7 k = 0$

$\Rightarrow k = - 2$

Question 9

The angle between the pair of straight lines $x^{2} - y^{2} - 2 y - 1 = 0$ , is

$90^{\circ}$

$60^{\circ}$

$75^{\circ}$

$36^{\circ}$

SOLUTION

Solution : A

Pair of straight lines represented by a second degree equation with coefficient of $x^{2}$ as a and coefficient of $y^{2}$ as b are perpendicular if a+b = 0
Here a + b = 0, so they are perpendicular to each other

Question 10

If the lines $(p - q) x^{2} + 2 (p + q) x y + (q - p) y^{2} = 0$ 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 $x^{2}$ as a and coefficient of $y^{2}$ 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 $x^{2} + y^{2} - 2 y + λ = 0$ to the origin are perpendicular, then the value of $λ$ will be

$\frac{1}{2}$

$\frac{- 1}{2}$

$\frac{1}{\sqrt{2}}$

SOLUTION

Solution : D

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

$x^{2} + y^{2} - 2 y (x + y) + λ (x + y)^{2} = 0$

$\Rightarrow x^{2} (1 + λ) + y^{2} (- 1 + λ) - 2 y x = 0$

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

$\Rightarrow 1 + λ - 1 + λ = 0 \Rightarrow λ = 0$

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 = $\pm$ 1. The locus of Q is

$12 x^{2} + 4 y^{2} = 3$

$12 x^{2} - 4 y^{2} = 3$

$12 x^{2} - 4 y^{2} + 3$ = 0

$12 x^{2} + 4 y^{2} + 3$ = 0

SOLUTION

Solution : B

According to the given condition

$\sqrt{(x - 1)^{2} + y^{2}} - \sqrt{(x + 1)^{2} + y^{2}}$ = $\pm$ 1

On squaring both sides, we get

$2 x^{2} + 2 y^{2} + 1 = 2 \sqrt{(x - 1)^{2} + y^{2}} \sqrt{(x + 1)^{2} + y^{2}}$

Again on squaring, we get $12 x^{2} - 4 y^{2} = 3$ .

Question 13

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

A. I quadrant

B. II quadrant

C. III quadrant

D. IV quadrant

SOLUTION

Solution : A

Coordinates of A and B are (-3, 4) and $(- \frac{3}{5}, \frac{8}{5})$ if orthocenter P(h, k)

Then, (slope of PA) $\times$ (slope of BC) = - 1
$\frac{k - 4}{h + 3} \times 4 = - 1$

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

and slope of PB $\times$ slope of AC = - 1
$\Rightarrow \frac{k - \frac{8}{5}}{h + \frac{3}{5}} \times - \frac{2}{3} = - 1$
$\Rightarrow \frac{5 k - 8}{5 h + 3} \times \frac{2}{3} = 1$
$\Rightarrow$ 10k - 16 = 15th + 9
15th - 10k + 25 = 10
3h - 2k + 5 = 0 ...(ii)
Solving Eqs. (i) and (ii), we get $h = \frac{3}{7}, k = \frac{22}{7}$
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

y^{2} = 2 x

y^{2} = 4 x

y = 2 x

x + y = 2 x y

SOLUTION

Solution : A

a = x - intercept, b = y - intercept
$2 h = a + b, k^{2} = a b$
$\frac{x}{a} + \frac{y}{b} = 1$ , substitute (1, 1)
$\frac{1}{a} + \frac{1}{b} = 1$
a + b = ab
$2 h = k^{2} \Rightarrow y^{2} = 2 x$

Question 15

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)

(\frac{2}{3}, \frac{4}{3})

(\frac{1}{3}, \frac{2}{3})

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).

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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