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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. .

Q1.  The equation of the base of an equilateral triangle is x+y=2 and the vertex is (2,-1),then the length of the side of the triangle is
•  √(3/2)/√(2/3)
•  √2
•  √(2/3)
•  √(3/2)
Solution
(c) Let p be the length of the perpendicular from the vertex (2,-1) to the base x+y=2, then p=|(2-1-2)/√(12+12 )| =1/√2 If a be the length of the side of triangle then, p=a sin⁡〖60°〗 ⇒ 1/√2=(a√3)/2 ⇒ a=√(2/3)
Q2.The equation y2-x2+2x-1=0 represents
•  A hyperbola
•  An ellipse
•  A pair of straight lines
•  A rectangular hyperbola
Solution
c) Given equation of curve is y2-x2+2x-1=0 Here, a=-1,b=1,c=-1,h=0,g=1,f=0 ∴ ∆=abc+2fgh-af2-bg2-ch2 =(-1)(1)(-1)+2(0)1(0)-0-1-0 =1-1=0 ∴ Given equation is equation of pair of straight lines.
Q3.  If the diagonals of a parallelogram ABCD are along the lines x+5y=7 and 10x-2y=9, then ABCD must be a
•  Rectangle
•  Square
•  Rhombus
Solution
d) Clearly, diagonals are perpendicular So, ABCD must be a rhombus

Q4. The pairs of straight lines ax2+2hxy-ay2=0
and hx2-2axy-hy2 are such that
•  One pair bisects the angle between the other pair
•  The lines of one pair are equally inclined to the lines of the other pair
•  The lines of each pair are perpendicular to other pair
•  All of these
Solution
d) The two pairs of lines are ax2+2hxy-ay2=0 …(i) hx2-2axy-hy2=0 …(ii) Clearly, these two equations represent two pairs of lines such that the lines in each pair are mutually perpendicular. The combined equation of the bisectors of the angles between the lines given in (i) is (x2-y2)/(a+a)=xy/h⇒hx2-2axy-hy2=0 Clearly it is same as (ii). Thus, each pair bisects the angle between the other pair.
Q5.The equation of straight line equally inclined to the axes and equidistance from the points (1,-2) and (3,4) is ax+by+c=0
, where
•  a=1,b=-1,c=3
•  a=1,b=-1,c=-3
•  a=1,b=1,c=-3
•  None of these
Solution
(c) The equation of straight line equally inclined to the axes is x/a+y/a=1 ⇒ x+y=a. Since, it is equidistant from the points (1,-2) and (3, 4), so perpendicular distances from these points on the line will be equal. ⇒ |(1-2-a)/√(12+12 )|=|(3+4-a)/√(12+12)| ⇒ (1+a)/√2=(7-a)/√2 ⇒ 2a=6 ⇒ a=3 ∴ Equation is x+y-3=0 But, given equation is ax+by+c=0 ∴a=1,b=1,c=-3
Q6. The equation y=±√3 x,y=1 are the sides of
•  An equilateral triangle
•  A right angled triangle
•  An isosceles triangle
•  An obtuse triangle
Solution
(a) Equation of OA is y=√3 x. Equation of OB is y=-√3 x and equation of AB is y=1 Its an equilateral triangle.
Q7.If A(2,-1) and B(6,5) are two points, then the ratio in which the foot of the perpendicular from (4,1) to AB divided
it, is
•  8:15
•  5:8
•  -5:8
•  -8:5
Solution
(b) Let P(4,1)and PD⊥AB. Equation of AB is 3x-2y-8=0 ∴ Equation of PD is 2x+3y-11=0 Let line AB is divided by PD in the ratio λ:1, then intersecting point D((6λ+2)/(λ+1),(5λ-1)/(λ+1)) lies on 2x+3y-11=0 ⇒ 2((6λ+2)/(λ+1))+3((5λ-1)/(λ+1))-11=0 ⇒ 16λ-10=0 ⇒ λ:1=5:8
Q8.Origin containing angle bisector of two lines L1≡a1x+b1y+c1=0 and L2≡a2x+b2y+c2=0 (where c1c2<0) is
•  (a1x+b1y+c1)/√(a12+b12 )=(a2x+b2y+c2)/√(a22+b22 )
•  a1x+b1y+c1)/√(a12+b12)=-(a2x+b2y+c2)/√(a22+b22 )
•  (a1x+b1y+c1)/(a12+b12 )=(a2x+b2y+c2)/(a22+b22 )
•  Deoends on the value of c1 and c2
Solution
b)
Q9. f the distance of any point (x,y) from the origin is defined as d(x,y)=max {|x|,|y|},d(x,y)=a, non-zero constant, then the locus is
•  A circle
•  A straight line
•  A square
•  A triangle
Solution
b) d(x,y)=max⁡{|x|,|y|} …(i) but d(x,y)=a …(ii) From, Eqs. (i) and (ii), a=〖max 〗⁡{|x|,|y|} If |x|>|y|, then a=|x| ∴ x=±a and if |y|>|x|, then a=|y| ∴y=±a Therefore, locus represents a straight line
Q10. Which of the following pair of straight lines intersect at right angle?
•   22x=y(x+2y)
•  (x+y)2=x(y+3 x)
•  2y(x+y)=xy
•  y=±2x
Solution
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Point and Straight Line Quiz-1
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