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, $\frac{1 + 3 + c}{3}$ = 0 $\Rightarrow$ 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 = $\frac{a + b + c}{3}$ , y = $\frac{b - c + c - a + a - b}{3}$ = 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 $x^{2} + y^{2} - 4 x + 6 y - 7 = 0$ 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

$x^{' 2} + y^{' 2} + x^{'} (2 h - 4) + y^{'} (2 k + 6) + h^{2} + k^{2} - 7 = 0$

To eliminate linear terms, we should have

2h - 4 = 0, 2k + 6 = 0 $\Rightarrow$ 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

4, \frac{8}{3}

3, 4

4, 3

- 3, 4

SOLUTION

Solution : A

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


Slope of BC = $\frac{6 - 0}{4 - 0} = \frac{3}{2}$

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

Slope of CA = $\frac{6 - 0}{4 - 0} = \frac{3}{2}$

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

(y – 0) = $- \frac{2}{3}$ (x – 8) i.e., 2x + 3y = 16                                                         …… (2)

Solving (1) and (2), the orthocenter is $4, \frac{8}{3}$

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

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
$t a n α = \frac{m - 7}{1 + 7 m} = \frac{m + 1}{1 + m (- 1)} \Rightarrow m = \frac{1}{3} o r - 3$
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

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 = $(\frac{6 + 7}{2}, \frac{- 1 + 3}{2}) = (\frac{13}{2}, 1) ∴ slope of P S = \frac{2 - 1}{2 - \frac{13}{2}} = - \frac{2}{9} ∴ The required equation is y + 1 = \frac{- 2}{9} (x - 1) i . e ., 2 x + 9 y + 7 = 0$

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

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 $\frac{x}{6} + \frac{y}{4} = 1$
$\Rightarrow 2 x + 3 y = 12$

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

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 $Δ A B C$ .

Now, area of $Δ A B C$
$= \frac{1}{2} | k - 1 | .1$
$\Rightarrow 1 = \frac{1}{2} | k - 1 |$
$\Rightarrow k - 1 = \pm 2$
$\Rightarrow k = - 1, 3$

Question 9

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

A. 3

¯ x = 2 α + u a n d 3 ¯ y = 2 β + v

B. 3

¯ x = 2 α - u a n d 3 ¯ y = 2 β - v

C. 3

¯ x = 2 α - u a n d 3 ¯ y = 2 β + v

D. 3

¯ x = 2 α + u a n d 3 ¯ y = 2 β - v

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,
$¯ x = \frac{2 α + u}{2 + 1} a n d ¯ y = \frac{2 β + v}{2 + 1}$
$\Rightarrow 3 ¯ x = 2 α + u a n d ¯ y = 2 β + v$

Question 10

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

$\frac{7}{3}$

$\frac{9}{2}$

$\frac{11}{2}$

$\frac{5}{4}$

SOLUTION

Solution : C

Distance between parallel lines $a x + b y + c_{1} = 0$ and $a x + b y + c_{2} = 0$ is $\frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}$

Given lines are $8 x + 6 y + 5 = 0$ and
$4 x + 3 y - 25 = 0$ or $8 x + 6 y - 50 = 0$

$∴$ Required distance = $\frac{5 + 50}{\sqrt{8^{2} + 6^{2}}} = \frac{55}{10} = \frac{11}{2}$

Question 11

The angle between the pair of straight lines $x^{2} + 4 y^{2} - 7 x y = 0,$ is

$t a n^{- 1} \frac{1}{3}$

$t a n^{- 1} 3$

$t a n^{- 1} \frac{\sqrt{33}}{5}$

$t a n^{- 1} \frac{5}{\sqrt{33}}$

SOLUTION

Solution : C

$tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$
$θ = t a n^{- 1} \frac{2 \sqrt{\frac{49}{4} - 4}}{5} = t a n^{- 1} \frac{\sqrt{33}}{5}$ .

Question 12

If (a + 3b)(3a + b) = $4 h^{2}$ , then the angle between the lines represented by $a x^{2} + 2 h x y + b y^{2} = 0$ is

$90^{\circ}$

$45^{\circ}$

$60^{\circ}$

${tan}^{- 1} \frac{1}{2}$

SOLUTION

Solution : C

$θ = t a n^{- 1} (\frac{2 \sqrt{h^{2} - a b}}{a + b}) = t a n^{- 1} (\frac{\sqrt{4 h^{2} - 4 a b}}{a + b})$
= $t a n^{- 1} (\frac{\sqrt{3 a^{2} + 3 b^{2} + 10 a b - 4 a b}}{a + b}) = 60^{\circ}$ .

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, $∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 3 & 4 & 1 x & y & 1 \end{matrix} ∣ ∣ ∣ ∣$ = 0 $\Rightarrow$ 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

$y^{2} = 4 a x$

$x^{2} = 4 a y$

$y^{2} + 4 a x = 0$

$x^{2} + 4 a y = 0$

SOLUTION

Solution : A

$(x - a)^{2} + y^{2} = (x + a)^{2} \Rightarrow y^{2} = 4 a x$

Note: This is also the definition of parabola $y^{2}$ = 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

$S Q^{2} + S R^{2} = 2 S P^{2}$ 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

$(x + 1)^{2} + y^{2} + (x - 2)^{2} + y^{2} = 2 [(x - 1)^{2} + y^{2}]$

$\Rightarrow$ 2x +1 + 4 - 4x = - 4x + 2 $\Rightarrow$ x = - $\frac{3}{2}$

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

Free Objective Test 02 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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