Geometry(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

GEOMETRY MCQs

Coordinate Geometry, Coordinate Geometry (10th Grade), Three Dimensional Geometry (10th Grade)

Total Questions : 133 | Page 6 of 14 pages

Question 51.

The lines 3x - 4y + 6 = 0 and 4x + 3y - 10 = 0 are mutually

parralel
perpendicular
none of these

Discuss Question

Answer: Option B. -> perpendicular

Question 52. The point C (–1,2) divides the line segment AB in the ratio 3:4, where the coordinates of A is (2, 5). The coordinates of B are___________.

(–5, –2)
(–1, – 2)
(–4, – 2)
(5, – 4)

Discuss Question

Answer: Option A. -> (–5, –2)
:
A
If a point

P (x, y)

divides a line segment joining

(x_{1}, y_{1}) and (x_{2}, y_{2})

in the ratio

m : n

, then the coordinates of P are given by:

x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n}

Let

C (- 1, 2)

dividethe line joining

A (2, 5)

and

B (x, y)

in the ratio 3:4.
Then,

C (\frac{3 x + 8}{7}, \frac{3 y + 20}{7})

= C(-1, 2)

\Rightarrow \frac{3 x + 8}{7} = - 1 3 x + 8 = - 7 \Rightarrow x = - 5 \Rightarrow \frac{3 y + 20}{7} = 2 3 y + 20 = 14 \Rightarrow y = - 2

Thus, thecoordinates of B are

B (- 5, - 2) .

Question 53. The points

(a, a), (- a, - a)

and

(- \sqrt{3} a, \sqrt{3} a)

are the vertices of a/an _____ triangle.

scalene
right angled
isosceles right angled
equilateral

Discuss Question

Answer: Option D. -> equilateral
:
D
Let's name the points as

A (a, a), B (- a, - a)

and

C (- \sqrt{3} a, \sqrt{3} a)

.
By distance formula, distance between points

(x_{1}, y_{1})

and

(x_{2}, y_{2})

= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

B C = \sqrt{(- \sqrt{3} a + a)^{2} + (\sqrt{3} a + a)^{2}} = \sqrt{a^{2} + (\sqrt{3} a)^{2} - 2 \times a \times \sqrt{3} a + \sqrt{3} a)^{2} + a^{2} + 2 \times a \times \sqrt{3} a} B C = \sqrt{8 a^{2}} = 2 \sqrt{2} a units Similarly A B = \sqrt{(a + a)^{2} + (a + a)^{2}} = \sqrt{8 a^{2}} = 2 \sqrt{2} a units A C = \sqrt{(- \sqrt{3} a - a)^{2} + (\sqrt{3} a - a)^{2}} = \sqrt{8 a^{2}} = 2 \sqrt{2} a units

Weget, A B = B C = A C = 2 \sqrt{2} a units ∴ △ ABC is an equilateral triangle .

Question 54. Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) is:

(– 7, 6)
(2, – 2)
(7, – 6)
(– 2, 2)

Discuss Question

Answer: Option B. -> (2, – 2)
:
B
Midpoint of theline segment joining

(x_{1}, y_{1})

and

(x_{2}, y_{2})

is given by

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

∴

Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) =

(\frac{- 5 + 9}{2}, \frac{4 - 8}{2})

= (2,-2)

Question 55. The co-ordinates of the points which divides the line joining (– 2, – 2) and (– 5, 7) in the ratio 2 : 1 is _____.

(4, – 4)
(– 3, 1)
(– 4, 4)
(1, – 3)

Discuss Question

Answer: Option C. -> (– 4, 4)
:
C
If a point P(x,y) divides the line joining

(x_{1}, y_{1}) and (x_{2}, y_{2})

in the ratio m:n
then

x

\frac{m x_{2} + n x_{1}}{m + n}

and y =

\frac{m y_{2} + n y_{1}}{m + n}

∴ x

\frac{m x_{2} + n x_{1}}{m + n}

\frac{2 (- 5) + 1 (- 2)}{2 + 1}

= -4
and

y

\frac{m y_{2} + n y_{1}}{m + n}

\frac{2 (7) + 1 (- 2)}{2 + 1}

= 4

∴

the required point is (-4,4).

Question 56. If the direction cosines of a variable line in two adjacent positions be l, m, n and l + a, m + b, n + c and the small angle between the two positions be

θ

, then :

θ=a+b+c
θ2=a2+b2+c2
|θ|=|a|+|b|+|c|
θ3=a3+b3+c3

Discuss Question

Answer: Option B. -> θ2=a2+b2+c2
:
B

l^{2} + m^{2} + n^{2} = 1, (l + a)^{2} + (m + b)^{2} + (n + c)^{2} = 1 \Rightarrow a l + b m + c n = - \frac{1}{2} (a^{2} + b^{2} + c^{2})

If The Direction Cosines Of A Variable Line In Two Adjacent ...

c o s θ = (a + l) l + m (b + m) + n (c + n)

= 1 + al + bm + cn

\Rightarrow a^{2} + b^{2} + c^{2} = 2 (1 - c o s θ) = 4 s i n^{2} (\frac{θ}{2})

Since

θ

is small,

s i n (\frac{θ}{2}) \approx \frac{θ}{2}

∴ θ^{2} = a^{2} + b^{2} + c^{2} [∵ s i n^{2} (\frac{θ}{2}) \approx \frac{θ^{2}}{4}]

Question 57. Direction cosines of the line which is perpendicular to the lines whose direction ratios are 1, –1, 2 and 2, 1, –1 are given by :

[−1√35,5√35,3√35]
[−1√35,−5√35,3√35]
[1√35,5√35,−3√35]
[1√35,−7√35,−3√35]

Discuss Question

Answer: Option A. -> [−1√35,5√35,3√35]
:
A
If l, m, n are the d.c.’s of the line which is perpendicular to the given lines, then
l – m + 2n = 0 and 2l + m –n = 0

∴ \frac{l}{1 - 2} = \frac{m}{4 + 1} = \frac{n}{1 + 2} \Rightarrow \frac{l}{- 1} = \frac{m}{5} = \frac{n}{3}

.
so d.r's =

(- 1, 5, 3)

Question 58. The equation of the plane passing through the points (1,-3,-2) and perpendicular to planes x +2y+2z=5 and 3x+3y+2z=8, is

2x -4y +3z-8 =0
2x-4y-3z+8 =0
2x+4y+3z+8 =0
x-3y+z-5=0

Discuss Question

Answer: Option A. -> 2x -4y +3z-8 =0
:
A
l+2m+2n=0, 3l+3m+2n=0,

l^{2} + m^{2} + n^{2} = 1

, we get l, m, nfrom these equations and then putting the values in l(x-1)+m(y+3) +n(z+2)=0, we get the required result.
Trick: Checking conversely,
2(x)-4(y)+3(z)-8 =0,
So, it passes through given point.
1(2)+2(-4)+2(3) =0,
So, it is perpendicular to x+2y+2z=5.
3(2)+3(-4)+2(3)=0,
So, it is perpendicular to 3x+3y +2z=8.

Question 59. If