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Point and Straight Line Quiz-6

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 line passing through (-1,Ï€/2) and perpendicular to √3 sin⁡Î¸ + 2cos⁡Î¸=4/r, is
•  2=√3rcos⁡Î¸ - 2rsin⁡Î¸
•  5=-2√3rsin⁡Î¸ + 4rcos⁡Î¸
•  2=√3rcos⁡Î¸ + 2r sin⁡Î¸
•  5=2√3 rsin⁡Î¸ - 4rcos⁡Î¸
Solution
a) Any line which is perpendicular to √3 sin⁡Î¸+2 cos⁡Î¸= 4/r is √3 sin⁡(Ï€/2+Î¸)+2 cos⁡(Ï€/2+Î¸)=k/r …(i) Since, it is passing through (-1,Ï€/2) ∴ √3 sin⁡Ï€+2 cos⁡Ï€=k/(-1)⇒k=2 On putting k=2 n Eq. (i), we get √3 cos⁡Î¸-2 sin⁡Î¸=2/r ⇒2=√3 r cos⁡Î¸-2r sin⁡Î¸
Q2.If x2/a+y2/b+2xy/h=0 represents a pair of straight lines such that slope of one line is twice the other, then ab∶h2 is
•  9∶8
•  8∶9
•  1∶2
•  2∶1
Solution
a)
Q3.  The length of the perpendicular from the origin of the line (x sin⁡Î±)/b - (y cos⁡Î±)/a - =0 is
•  (|ab|)/√(a2cos2Î±-b2sin2Î±)
•  (|ab|)/√(a2cos2Î±+b2sin2Î±)
•  (|ab|)/√(a2sin2Î±-b2cos2Î±)
•  (|ab|)/√(a2 sin2 Î±+b2cos2Î±)
Solution
(d) The length of perpendicular from the origin to the line (x sin⁡Î±)/b-(y cos⁡Î±)/a-1=0 is d=|0-0-1|/√((sin2Î±)/b2 +(cos2 Î±)/a2 ) =(|ab|)/√(a2sin2Î±+b2cos2Î±)

Q4. Let the perpendiculars from any point on the line 2x+11y=5 upon the lines 24x+7y-20=0 and 4x-3y-2=0 have the lengths p1 and p2 respectively. Then,
•  2p1=p2
•  p1=p2
•  p1=2p2
•  None of these
Solution
b) Let (t,(5-2t)/11) be a point on the line 2x+11y=5 Then, p1=|(24t+7((5-2t)/11)-20)/√(242+72)|=(|50t-37|)/55 and, p2=|(4t-3((5-2t)/11)-2)/√(42+(-3)2)|=(|50t-37|)/55 Clearly, we have p1=p2
Q5.Two consecutive sides of a parallelogram are 4x+5y=0 and 7x+2y=0. One diagonal of the parallelogram is 11x+7y=9. If
the other diagonal is ax+by+c=0, then
•  a=-1,b=-1,c=2
•  a=1,b=-1,c=0
•  a=-1,b=-1,c=0
•  a=1,b=1,c=0
Solution
(b) Since, the coordinates of three vertices A,B and C are (5/3,-4/3),(0,0) and (-2/3,7/3) respectively, also the mid point of AC is (1/2,1/2), therefore the equation of line passing through (1/2,1/2) and (0, 0) is given by x-y=0, which is the required equation of another diagonal, so a=1,b=-1,and c=0
Q6. The value of Î», for which the equation x2-y2-x+Î»y-2=0 represents a pair of straight lines, are
•  -3,1
•  -1,1
•  3,-3
•  3,1
Solution
(c) On comparing given equation with standard equation, we get a=1,b=-1,c=-2,h=0,g=-1/2 ,f=Î»/2 Given equation represent a pair of straight line, ∴abc+2 fgh-af2-bg2-ch2=0 ⇒ 2+0-Î»2/4+1/4=0 ⇒ Î»2/4=9/4 ⇒ Î»=±3
Q7.The equations ax+by+c=0 and dx+ey+f=0 represent the same straight line if and only if
•  a/d = b/c
•  c = f
•  a/d = b/e = c/f
•  a = d,b = e,c = f
Solution
c) The equations ax+by+c=0 and dx+ey+f=0 will represent the same straight line if their slopes and y-intercepts are equal ∴-a/b=-d/e and-c/b=-f/e ⇒a/d=b/e and b/e=c/f⇒a/d=b/e=c/f
Q8.If the angle Î¸ is acute, then the acute angle between x2(cos⁡Î¸-sin⁡Î¸ )+2xycos⁡Î¸+y2(cos⁡Î¸+sin⁡Î¸)=0 is
•  2 Î¸
•  Î¸/3
•  Î¸
•  Î¸/2
Solution
c) On comparing the given equation with standard equation, we get a=cos⁡Î¸-sin⁡Î¸,b=cos⁡Î¸+sin⁡Î¸,h=cos⁡Î¸ tan⁡Ï•=(2√(cos2 Î¸-(cos2Î¸-sin2Î¸) ))/cos⁡Î¸-sin⁡Î¸+cos⁡Î¸+sin⁡Î¸ =(2 sin⁡Î¸)/(2 cos⁡Î¸ ) ⇒tan⁡Ï•=tan⁡Î¸⇒Ï•=Î¸
Q9.  The ratio in which the line 3x-2y+5=0 divides the join of (6,-7) and (-2,3), is
•  1∶1
•  7∶37
•  37∶7
•  None of these
Solution
c)
Q10. A straight line through P(1,2) is such that its intercept between the axes is bisected at P. Its equation is
•  x+2 y=5
•   x-y+1=0
•  x+y-3=0
•  2x+y-4=0
Solution
d) Let the equation of the line be x/a+y/b=1. This cuts the coordinates axes at A(a,0) and B(0,b) The coordinates of the mid-point of the intercept AB between the axes are (a/2,b/2) ∴a/2=1,b/2=2 ⇒a=2,b=4 Hence, the equation of the line is x/2+y/4=1 or, 2 x+y=4

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