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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 equations a2x2+2h(a+b)xy+b2y2=0 and ax2+2hxy+by2=0 represent
•  Two pairs of perpendicular straight lines
•  Two pairs of parallel straight lines
•  Two pairs of straight lines which are equally inclined to each other
•  None of these
Solution
(c) We have, a2x2+2h(a+b)xy+b2y2=0 …(i) ax2+2hxy+by2=0 …(ii) The equation of the bisectors of the angles between the pair of lines given in (i) is (x2-y2)/(a2-b2 )=xy/(h(a+b))⇒(x2-y2)/(a-b)=xy/h This is same as the equation of the bisectors of the angles between the lines given in (ii). Thus, two pairs of straight lines are equally inclined to each other
Q2.If the points (1,2) and (3,4) were to be on the same side of the line 3x-5y+a=0, then
•  7
•  a=7
•  a=1
•   a<7 or a>11
Solution
d) If the points (1,2) and (3,4) are on the same side of 3x-5y+a=0, then (3-10+a) and 9-20+a are of the same sign ∴(3-10+a)(9-20+a)>0 ⇒(a-7)(a-11)>0⇒a<7 or a>11
Q3.  The sum of slopes of lines 3x2+5xy-2y2=0 is
•  -5/3
•  5/2
•  -5/2
•  -2/3
Solution

b) Sum of slope of the lines 3x2+5xy-2y2=0 is m1+m2=-h/b=5/2

Q4. If (-4,5) is one vertex and 7x-y+8=0 is one diagonal of a square, then the equation of the second diagonal is
•  a) x+3y=21
•  2x-3y=7
•  x+7y=31
•  2x+3y=21
Solution
(c) Equation of perpendicular diagonal to 7x-y+8=0 is x+7y=λ, which passes through (-4,5) ∴ λ=31 So, equation of another diagonal is x+7y=31
Q5.In a rhombus ABCD the diagonals AC and BD intersect at the point (3,4). If the point A is (1,2) the diagonal BD has
the equation
•  x-y-1=0
•  x+y-1=0
•  x-y+1=0
•  x+y-7=0
Solution
d) Since the diagonals of a rhombus bisect each other at right angle. Therefore, BD passes through (3,4) and is perpendicular to AC. So, its equation is y-4=-1(x-3)⇒x+y-7=0
Q6.  The equation of one of the lines parallel to 4x-3y=5 and at a unit distance from
the point (-1,-4) is
•  3x+4y-3=0
•   3x+4y+3=0
•  4x-3y+3=0
•  4x-3y-3=0
Solution
(d) Required equation can be 4x-3y-K=0 ∴|(4×-1-3×-4-K)/√(42+(-3)2 )|=1 ⇒(-4+12-K)/5=±1 ⇒8-K=±5 ⇒K=3 or K=13 ∴ Equation of lines are 4x-3y-3=0 and 4x-3y-13=0
Q7.A point moves in such a way that the square of its distance from the point (3,-2) is equal to numerically its distance
from the line 5x-12y=13. The equation of the locus of the point is
•  x2+y2-11x-16y+26=0
•  x2+y2-11x+16y=0
•  13(x2+y2)-83x+64y+182=0
•  x2+y2-83x+64y+182=0
Solution
(c) Let (h,k ) be the point such that (h-3)2+(k+2)2=(5h-12k-13)/√(25+144) ⇒13(h2+9-6h+k2+4k+4)=5h-12k-13 ⇒13(h2+k2 )-83h+64k+182=0 Thus, the locus of (h,k) is 13(x2+y2)-83x+64y+182=0
Q8.The centroid of the triangle whose three sides are given by the combined equation (x2+7xy+2y2)(y-1)=0, is
•  (2/3,0)
•  (7/3,2/3)
•  (-7/3,2/3)
•  None of these
Solution

(c) The sides of the triangle are y=1 and the pair of lines x2+7 xy+2y2=0 Clearly, one vertex is (0, 0) and the y-coordinates of each of the other two vertices is 1. On putting y=1 in the second equation, we get x2+ 7x+2=0 If x1 and x2 are the roots of this equation, then x1+x2=-7 ∴Centroid,G=((0+x1+x2)/3,(0+1+1)/3) =(-7/3,2/3)
Q9.  If the equation 12 x2+7xy-py2-18 x+qy+6=0 represents a pair of perpendicular straight lines, then
•  p=12,q=1
•  p=1,q=12
•  p=-1,q=12
•  p=1,q=-12
Solution
(a) The equation 12 x2+7xy-py2-18x+qy+6=0 will represent a pair of perpendicular lines -72 p-63/2 q-3 q2+81 p-147/2=0 and 12-p=0 ⇒2 q2+21 q-23=0 and p=12 ⇒q=1and p=12
Q10. If P(sin⁡θ,1/√2) and Q(1/√2,cos⁡θ), -π≤θ≤π are two points on the same side of the line x-y=0, then θ belongs to the interval
•  -π/4,π/4
•   (-π/4,π/4)
•  (π/4,3 π/4)
•  None of these
Solution
(a) If P(sin⁡θ,1/√2) and Q(1/√2,cos⁡θ) are on the same side of the line x-y=0. Then, ⇒(sin⁡θ-1/√2 )(1/√2-cos⁡θ )>0 ⇒(sin⁡θ-1/√2)(cos⁡θ-1/√2)<0 ⇒sin⁡θ-1/√2>0 and cos⁡θ-1/√2<0 or,sin⁡θ-1/√2<0 and cos⁡〖θ-1/√2>0 ⇒(sin⁡θ>1/√2 and cos⁡θ<1/√2) or,(sin⁡θ<1/√2 and cos⁡θ>1/√2 ) ⇒θ∈(π/4,3 π/4) or, θ∈(-π/4,π/4) ⇒θ∈(-π/4,π/4)∪(π/4,3 π/4) #### Written by: AUTHORNAME

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: Point and Straight Line Quiz-10
Point and Straight Line Quiz-10
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