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Hinduja

Variable straight lines L1 =2x+C1 and L2=2x+C2 meet the x-axis in A1 and A2 respectively and y-axis in B1 and B2 respectively. Locus of intersection point of A1B2 and A2B1 is

a)y+x=0

b)y=x

c)y+2x=0

d)y=2x

The no. of real values of k for wich the lines x-2y+3=0 ,kx+3y+1=0 and 4x-ky+2=0 are concurrent.

a)0

b)1

c)2

d)infinite

The no. of integral points exactly in the interior of the triangle with vertices (0,0) (0,21) (21,0) is

a) 133

b) 190

c) 233

d) 105

nitin tewari

ans 1)c

ans 2)a

ans 3)b

soln:

1)form the equation for line joining a1to b2 and a2 to b1

on solving for the point of intersection u get the point of intersection as

x= -(c1²c2- c1c2²)/(2c1²-2c2²)

and

y=2(c1²c2- c1c2²)/(2c1²-2c2²)

so u get y=-2x i.e y+2x =0

2)for concurrency condition u should have the determinant formed by putting the coefficients and costant term should be zero.

from there we get a quad. in k for which D(discriminant comes out to be -ve) so no solution for k.

3)for exactly inside lying condition u see the points lying on lines parallel to y axis.

on x=1 u get 20-1 such points

on x=2 u get 20-2 such points

on x=3 u get 20-3 points.

........

so all u require is

_{[ 1]}^{[19 ]} n = 19*20/2=190

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Bipin Dubey

The points are the intersection of the lines with the axes :

A₁ is (-c₁/2,0) and A₂ is (-c₂/2,0).
B₁ is (0,c₁) and B₂ is (0,c₂).

Now find equations of A₁B₂ and A₂B₁ and find their intersection.

You will get : { -(c₁²c₂ + c₂²c₁)/2(c₁²-c₂²) , (c₁²c₂ + c₂²c₁)/(c₁²-c₂²)}

Clearly its locus is y+2x = 0.

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