Point and Straight Line Quiz-5

Point and Straight Line Quiz-5

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. The coordinates of three vertices of a quadrilateral in order are (6,1),(7,2) and (-1,0). If the area of the quadrilateral is 4 square units, then the locus of the fourth vertex is

•  x-7y=1
•  x-7y+15=0
•  (x-7y)²+14(x-7y)-15=0
•  None of these

Solution
c) Let (h,k) be the coordinates of the fourth vertex. Then, ∆₁ =1, ∆₃=1/2(h-6k) We have, |∆_{1,/sub>+∆2+∆3+∆4|=4 ⇒|5/2+1-k/2+(h-6k)/2|=4 ⇒|7+h-7k|=8 ⇒7+h-7k=±8 ⇒h-7k-1=0,h-7k+15=0 ⇒(h-7k-1)(h-7k+15)=0 ⇒(h-7k)²+14(h-7k)-15=0 Hence, the locus of (h,k) is (x-7y)²+14(x-7y)-15=0}

Q2.The value of k such that 3x²-11 xy+10y²-7x+13y+k=0 may represent a pair of straight lines, is

•  3
•  4
•  6
•  8

Solution
b)

Q3.  A straight line through P(1,2) is such that its intercept between the axes is bisected at P. Its equation is

•  x+y=-1
•   x+y=3
•  x+2y=5
•  2x+y=4

Solution
(d) ∵ P(1,2) is mid point of AB, therefore coordinate of A and B respectively (2, 0) and (0,4). ∴ Equation of line AB is y-0=4/(-2) (x-2) ⇒ 2x+y=4

Q4. In the above question the coordinates of the other two vertices are

•  (2,0),(4,4)
•  (2,4),(4,0)
•  (-2,0),(4,-4)
•  (2,0),(-4,4)

Solution
(a) Let (x,y) be the coordinates of the vertex B. Then, BE=1/2 AC ⇒(x-3)^{2+(y-2)2=((1-5)2+(3-1)2)/4 ⇒(x-3)2+(y-2)=5 …(i) Solving (i) with y=2 x-4, we get coordinates of B and D as (2,0) and (4,4) respectively}

Q5.If the lines x²+2xy-35y²-4x+44y-12=0 and 5x+λy-8=0 are concurrent, then the value of λ is

•  0
•  1
•  -1
•  2

Solution
  d) Given line is x²+2xy-35y²-4x+44y-12=0
, a=1,b=-35,c=-12,h=1,f=22
Point of intersection=((22-70)/(-35-1),(-2-22)/(-35-1))=(4/3,2/3)
Ifthe lines are concurrent. The point (4/3,2/3) will be on the line 5x+λy-8=0
∴5(4/3)+λ(2/3)-8=0
⇒ 2/3 λ=8-20/3=4/3⇒λ=2

Q6. A point equidistant from the lines 4 x+3 y+10=0,5 x-12 y+26=0 and 7 x+24 y-50=0 is

•  (1,-1)
•  (1,1)
•  (0,0)
•  (0,1)

Solution
c) Clearly, lengths of perpendiculars from (0,0) on the gives lines are each equal to 2. Hence, required point is (0,0)

Q7.If the lines x+3y-9=0,4x+by-2=0 and 2x-y-4=0 are concurrent, then b equals

•  -5
•  5
•  1
•  0

Solution
a) Given lines are concurrent, then ⇒ 1(-4b-2)-3(-16+4)-9(-4-2b)=0 ⇒ 14b+70=0 ⇒ b=-5

Q8.Consider the following statements:
I. The lines 2x+3y+19=0 and 9x+6y-17=0 cut the coordinates axes in concyclic points
II.The points (2,-5) and (-1,4) are equidistant from the line 3x+y+5=0
Which of these is are correct?

•  Only (1)
•  Only (2)
•  Both of these
•  None of these

Solution
(c) (1)Let A and B be the points where the lines 2x+3y+19=0 meets the coordinates axes and let C and O be the points where the line 9x+6y-17=0 meet
coordinate axes Then,OA=19/2,OB=19/2,
OC=17/9 and OD=17/6
Thus, the segments AOC and BOD intersect at such that OA∙OC=OB∙OD. Hence, A,B,C,D are concyclic (2) Distance of (2,-5) from the line 3x+y+5-0 is
(2×3-5+5)/√(3²+1²)=6/√10
and distance of (-1,4) from the line 3x+y+5=0 is (3(-1)+4+5)/√10=6/√10
Thus, the points are equidistant from the given line
Hence, both of these statements are correct

Q9.  49. If a line passes through the point (2,2) and encloses a triangle of area A square units with the coordinate axes, then the intercepts made by the line on the coordinate axes are the roots of the equations

•   x²±Ax∓2A=0
•  x²±Ax±2A=0
•  x²±2Ax±A=0
•  x²±2Ax∓A=0

Solution
a) Let the equation of the line be x/a+y/b=1 It passes through (2,2) ∴2/a+2/b=1⇒2(a+b)=ab …(i) The line encloses a triangle of area A square units with the coordinate axes ∴1/2 |a||b|=A⇒|ab|=2A⇒ab=±2A …(ii) From (i) and (ii), we get a+b=±A The quadratic equation having a,b as its roots is x²-x(a+b)+ab=0 or, x²∓Ax±2A=0

Q10. Consider the fourteen lines in the plane given by y=x+r,y=-x+r, where r∈{0,1,2,3,4,5,6}. The number of squares formed by these lines, whose sides are of length √2, is

•   9
•  16
•  25
•  36

Solution
c)