The Point (a2, a + 1) Is A Point In The Angle Between The Lines 3x – y + 1 = 0 And x + 2y – 5 = 0 Containing Origin. Then ‘a’ belongs To The Interval. 

Why Kaysons ?

Video lectures

Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.

Online Support

Practice over 30000+ questions starting from basic level to JEE advance level.

Live Doubt Clearing Session

Ask your doubts live everyday Join our live doubt clearing session conducted by our experts.

National Mock Tests

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

Organized Learning

Proper planning to complete syllabus is the key to get a decent rank in JEE.

Test Series/Daily assignments

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.



The point (a2a + 1) is a point in the angle between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0 containing origin. Then ‘a’ belongs to the interval. 


Correct option is

Point (a2a + 1) and origin lie on the same side of each straight line 



combining (i) and (ii) we get, 



If A (cos α, sin α), B (sin α, –cos α), C (1, 2) are the vertices of a ΔABC, then as α varies the locus of its centroid is:


The image of point (1, 3) in the line x + y – 6 = 0 is: 


If t1t2t3 are distinct, then the points  are collinear if:


A and B are two fixed points. The vertex C of a Δ ABC moves such that cot A + cot B = constant. Locus of C is a straight line:


The number of integer values of m, for which the x-co-ordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer, is:


The vertices of a triangles are A (x1x1 tan α), B (x2x2 tan β) and C (x3,x3 tan γ). If the circumcentre of Δ ABC coincides with the origin and H (ab) be its orthocenter, then a/b is equal to:


A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinates axes at points P and Q. As L varies, the absolute minimum values of OP + OQ is (O is origin)




Consider the family of lines 

           (x + y – 1) + λ (2x + 3y – 5) = 0

and   (3x + 2y – 4) + µ (x + 2y – 6) = 0  

equation of a straight line that belongs to both the families is:


If p1p2p3 be the length perpendicular from the points

(m2, 2m), (mm’m + m’) and (m’2, 2m’)

respectively on the line  



Find all points on x + y = 4 that lie at a unit distance from the line 4x + 3y– 10 = 0.