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.

.

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 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

Solution

c) Let (h,k) be the coordinates of the fourth vertex. Then, ∆

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 3x

^{2}-11 xy+10y

^{2}-7x+13y+k=0 may represent a pair of straight lines, is

Solution

b)

b)

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

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

(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

Solution

(a) Let (x,y) be the coordinates of the vertex B. Then, BE=1/2 AC ⇒(x-3)

(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 x

^{2}+2xy-35y

^{2}-4x+44y-12=0 and 5x+Î»y-8=0 are concurrent, then the value of Î» is

Solution

d) Given line is x

, 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+Î»y-8=0

∴5(4/3)+Î»(2/3)-8=0

⇒ 2/3 Î»=8-20/3=4/3⇒Î»=2

d) Given line is x

^{2}+2xy-35y^{2}-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+Î»y-8=0

∴5(4/3)+Î»(2/3)-8=0

⇒ 2/3 Î»=8-20/3=4/3⇒Î»=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

Solution

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

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

Solution

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

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?

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)/√(3

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

(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)/√(3

^{2}+1^{2})=6/√10and 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

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 x

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 x

^{2}-x(a+b)+ab=0 or, x^{2}∓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

Solution

c)

c)