1
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
The region represented by| x – y | $$\le$$ 2 and | x + y| $$\le$$ 2 is bounded by a :
A
rhombus of area 8$$\sqrt 2$$ sq. units
B
square of side length 2$$\sqrt 2$$ units
C
square of area 16 sq. units
D
rhombus of side length 2 units
2
JEE Main 2019 (Online) 9th April Evening Slot
+4
-1
If the two lines x + (a – 1) y = 1 and 2x + a2y = 1 (a$$\in$$R – {0, 1}) are perpendicular, then the distance of their point of intersection from the origin is :
A
$${2 \over \sqrt5}$$
B
$${\sqrt2 \over 5}$$
C
$${2 \over 5}$$
D
$$\sqrt{2 \over 5}$$
3
JEE Main 2019 (Online) 9th April Morning Slot
+4
-1
Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is :
A
$${{\sqrt 7 - 1} \over {\sqrt 7 + 1}}$$
B
$${{\sqrt 5 - 1} \over {\sqrt 5 + 1}}$$
C
$${{1 - \sqrt 5 } \over {1 + \sqrt 5 }}$$
D
$${{1 - \sqrt 7 } \over {1 + \sqrt 7 }}$$
4
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
If the system of linear equations

x – 2y + kz = 1
2x + y + z = 2
3x – y – kz = 3

has a solution (x,y,z), z $$\ne$$ 0, then (x,y) lies on the straight line whose equation is :
A
4x – 3y – 4 = 0
B
3x – 4y – 1 = 0
C
4x – 3y – 1 = 0
D
3x – 4y – 4 = 0
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