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12th Grade > Mathematics

STRAIGHT LINES MCQs

Straight Lines

Total Questions : 60 | Page 4 of 6 pages
Question 31. 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
  1.    a + c = -4
  2.    a + b = -4
  3.    c = -4
  4.    b + c = -4
 Discuss Question
Answer: Option C. -> c = -4
:
C
The point lies on axis of x, if y = 0.
Therefore, 1+3+c3 = 0 ⇒ c = -4.
Question 32. 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
  1.    2x + 3y = 12
  2.    3x+2y=12
  3.    4x - 3y = 6
  4.    5x - 2y = 10
 Discuss Question
Answer: Option A. -> 2x + 3y = 12
:
A
Using midpoint formula we get the coordinates of the points A and B as (6, 0) and (0, 4) respectively.
If The Co-ordinates Of The Middle Point Of The Portion Of A ...
Therefore, using intercept form of the striaght line, the equation of line AB is x6+y4=1
⇒2x+3y=12
Question 33. If (a + 3b)(3a + b) = 4h2, then the angle between the lines represented by ax2+2hxy+by2=0 is
  1.    90∘
  2.    45∘
  3.    60∘
  4.    tan−112
 Discuss Question
Answer: Option C. -> 60∘
:
C
θ=tan−1(2√h2−aba+b)=tan−1(√4h2−4aba+b)
= tan−1(√3a2+3b2+10ab−4aba+b)=60∘.
Question 34. 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
  1.    {1, 3}
  2.    {0, 2}
  3.    {-1, 3}
  4.    {-3, -2}
 Discuss Question
Answer: Option C. -> {-1, 3}
:
C
Since, A(h, k), B(1, 1) and C (2, 1) are the vertices of a right angled ΔABC.
Let A (h, K), B (1, 1) And C (2, 1) Be The Vertices Of A Rig...
Now, area of ΔABC
=12|k−1|.1
⇒1=12|k−1|
⇒k−1=±2
⇒k=−1,3
Question 35. 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
  1.    (3, 2)
  2.    (- 3, 2)
  3.    (2, - 3)
  4.    (1.7)
 Discuss Question
Answer: Option C. -> (2, - 3)
:
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 36. The angle between the pair of straight lines x2+4y2−7xy=0, is 
  1.    tan−113
  2.    tan−13
  3.    tan−1√335
  4.    tan−15√33
 Discuss Question
Answer: Option C. -> tan−1√335
:
C
tanθ=2√h2−aba+b
θ=tan−12√494−45=tan−1√335.
Question 37. If PM is the perpendicular from P (2,3) on to the line x+y=3 then the co-ordinates of M are
  1.    (2,1)
  2.    (−1,4)
  3.    (1,2)
  4.    (4,−1)
 Discuss Question
Answer: Option C. -> (1,2)
:
C
If(x,y) is the foot of the perpendicular from (x1,y1) to the line ax + by + c = 0 then
x−x1a=y−y1b=−(ax1+by1+c)a2+b2
Here (x1,y1)=(2,3);ax+by+c=x+y−3
∴ x−21=y−31=−((1×2)+(1×3)+(−3))12+12
⇒ x−21=y−31=−((2)+(3)+(−3))2
⇒ x−21=y−31=−22
⇒ x−21=y−31=−1
∴ x - 2 = -1 ; y-3 = -1
⇒ x = -1 + 2 ; y = -1 + 3
⇒ x = 1 ; y = 2
∴ (x,y) = (1,2)
Question 38. Two points A and B have coordinates (1, 0) and (-1, 0) respectively and Q is a point which satisfies the relation
AQ - BQ = ± 1. The locus of Q is
  1.    12x2+4y2=3
  2.    12x2−4y2=3
  3.    12x2−4y2+3 = 0
  4.    12x2+4y2+3 = 0
 Discuss Question
Answer: Option B. -> 12x2−4y2=3
:
B
According to the given condition
√(x−1)2+y2−√(x+1)2+y2 = ±1
On squaring both sides, we get
2x2+2y2+1=2√(x−1)2+y2√(x+1)2+y2
Again on squaring, we get 12x2−4y2=3.
Question 39. A line through the point A(2, 0), which makes an angle of 30∘ with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15∘. The equation of the straight line in the new position is
  1.    (2−√3)x−y−4+2√3=0
  2.    (2−√3)x+y−4+2√3=0
  3.    (2−√3)x−y+4+2√3=0
  4.    (2−√3)x−9y+4+2√3=0
 Discuss Question
Answer: Option A. -> (2−√3)x−y−4+2√3=0
:
A
Let AB be the initial position of the line and AC be its new position.
Slope of the line
AC=tan15∘=(2−√3)
∴ Equation of the line AC is
(y−0)=(2−√3)(x−2)ory=(2−√3)x−4+2√3or(2−√3)x−y−4+2√3=0
A Line Through The Point A(2, 0), Which Makes An Angle Of 30...
Question 40. If 4a+5b+6c=0 then the set of lines ax+by+c=0 are concurrent at the point
  1.    (23,56)
  2.    (13,12)
  3.    (12,43)
  4.    (13,73)
 Discuss Question
Answer: Option A. -> (23,56)
:
A
4a+5b+6c=0
(6x)a+(6y)b+6c=0 →[Multiply given set of lines ax+by+c=0 with '6']
Now on comparing 6x=4 and 6y =5
⇒ (x,y) = (23,56)
∴ ax+by+c=0 must passes through(23,56)

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