Point and Straight Line Quiz-8

Point and Straight Line Quiz-8

Dear Readers,

Dear Readers, As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.
Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
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Q1. If the lines kx-2y-1=0 and 6x-4y-m=2 are identical (coincident ) lines, then the values of k and m are

•  k=3,m=2
•  k=-3,m=2
•  k=-3,m=-2
•  k=3,m=-2

Solution
(a) Given lines are kx-2y-1=0 and 6x-4y-m=0 Since, these lines are coincident. ∴ k/6=(-2)/(-4)=(-1)/(-m) ⇒k/6=1/2 and 1/m=1/2 ⇒k=3 and m=2

Q2.The ratio in which the line 3 x+4 y+2=0 divides the distance between 3 x+4 y+5=0, and 3 x+4 y-5=0, is

•  7∶3
•  3∶7
•  2∶3
•  None of these

Solution
(b) Lines 3 x+4 y+2=0 and 3 x+4 y+5=0 are on the same side of the origin. The distance d₁ between these lines is given by
d₁=|(2-5)/√(3²+4² )|=3/5
Lines 3 x+4 y+2=0 and 3 x+4 y-5=0 are on the opposite sides of the origin. The distance d₂ between these lines is given by
d₂=|(2+5)/√(3²+4² )|=7/5
Thus, 3 x+4 y=0 divides the distance between 3x+4y+5=0 and 3 x+4 y-5=0 in the ratio d₁ ∶d₂ i.e. 3∶7

Q3.  The locus of the orthocentre of the triangle formed by the lines (1+p)x-py+p(1+p)=0,(1+q)x-qy+q(1+q)=0 and y=0, where
p≠q is

•  A hyperbola
•  A parabola
•  An ellipse
•  A straight line

Solution

d)

Q4. The locus of the point of intersection of lines x cos⁡〖α+y sin⁡α=a and x sin⁡α- ycos⁡α=b is (α is a variable)

•  2(x²+y²)=a²+b²
•  x²-y²=a²-b²
•  x²+y²=a²+b²
•  None of these

Solution
(c) Let (h,k) be the point of intersection of the line x cos⁡α+y sin⁡α=a and x sin⁡α-y cos⁡α=b. Then,
h cos⁡α+k sin⁡α=a …(i)
h sin⁡α-k cos⁡α=b …(ii)
Squaring and adding (i) and (ii), we get
(h cos⁡α+k sin⁡α)²+(h sin⁡α-k cos⁡α)²=a²+b²
⇒h²+k²=a²+b²
Hence, locus of (h,k) is x²+y²=a²+b²

Q5.If the equation 3x²+xy-y²-3x+6y+k=0 represents a pair of straight lines, then the values of k is

•  9
•  1
•  -9
•  0

Solution
  (c)
We have, 3x²+xy-y²-3x+6y+k=0 …(i)
Comparing this equation with
ax²+2hxy+by²+2gx+2fy+c=0, we have
a=3,b=-1,h=1/2,c=k,f=3 and g=-3/2
Equation (i) will represent a pair of straight lines if
abc+2fgh-af²-bg²-ch²=0
⇒-3k-9/2-27+9/4-k/2=0
⇒-13k/3-117/4=0⇒k=-9

Q6. The equation 2x²-24xy+11y²=0 represents

•  Two parallel lines
•  Two perpendicular lines
•  Two lines passing through the origin
•  A circle

Solution
a) The required lines are obtaining by shifting the origin at (4,0). So, the required equation is y=|x-4|

Q7.If A(cos⁡α, sin⁡α), B(sin⁡α,-cos⁡α), C(1,2) are the vertices of a ∆ ABC, then as α varies the locus of its centroid is

•  x²+y²-2 x-4 y+1=0
•  3(x²+y²)-2 x-4 y+1=0
•  x²+y²-2 x-4 y+3=0
•  None of these

Solution

b) Let (h,k) be the centroid of the triangle having vertices A(cos⁡α,-cos⁡α) and C(1,2). Then,
h=cos⁡α+sin⁡α+1/3 and k=sin⁡α-cos⁡α+2/3
⇒3 h-1=cos⁡α+sin⁡α and 3 k-2=sin⁡α-cos⁡α
⇒(3 h-1)²+(3 k-2)²=2 [Squaring and adding]
⇒9(h²+k²)-6 h-12 k+3=0
⇒3(h²+k²)-2 h-4k+1=0
Hence, the locus of (h, k) is 3(x²+y²)-2 x-4 y+1=0

Q8.The number of points on the line 3x+4y=5, which are at a distance of sec²θ + 2cossec²θ,θ∈R, from the point (1, 3) is

•  1
•  2
•  3
•  Infinite

Solution

b) The perpendicular distance of (1, 3) from the line 3x+4y=5 is 2 units while, sec²θ+2 cosec²θ≥3 [as sec²θ,cosec²θ≥1] So, there will be two such points on the line

Q9.  If the pair of straight lines given by Ax²+2Hxy+By²=0(H²>AB) forms an equilateral triangle with line ax+by+c=0, then (A+3B)(3A+B) is equal to

•  H²
•  -H²
•  2H²
•  4H²

Solution
(d) Given, Ax²+2Hxy+By²=0 …(i) and ax+by+c=0 …(ii) Since, triangle is equilateral, then angle between the two lines is 60° Angle between pair of lines is given by cos⁡60°=(A+B)/√((A-B)²+4H²) ⇒ (A+B)/√((A-B)²+4H²)=1/2 ⇒ (A-B)²+4H²=4(A+B)² ⇒ 4(A²+B²+2AB)-(A²+B²-2AB)=4H² ⇒ 3A²+10AB+3B²=4H² ⇒ (3A+B)(A+3B)=4H²

Q10. The equation of pair of lines joining origin to the points of intersection of x²+y²=9 and x+y=3 is

•   x²+(3-x)²=9
•   xy=0
•   (3+y)²+y²=9
•  (x-y)²=9

Solution
(b) Given, x²+y²=9 …(i)
and x+y=3 …(ii)
From Eqs. (i) and (ii), we make a homogeneous equation.
⇒ x²+y²=(x+y)²
⇒ x²+y²=x²+y²+2xy
⇒ xy=0