Straight Lines-Quiz-3

Straight Lines Quiz-3

Dear Readers,

The unit Straight Line holds sheer significance in the JEE Advanced and other engineering exams. It has a weightage of 6% pairing with Circles. With focused practice good marks can be fetched from this section. These questions are important in achieving your success in JEE and Other Engineering Exams..

Q1. If the equations y=mx+c and x cos⁡α+y sin⁡α=p represent the same straight line, then

•  p=c√(1+m² )
•  c=p√(1+m² )
•  cp=√(1+m² )
•  p² + c² + m² = 1

Solution

Q2.The graph y²+2xy+40|x|=400 divides the plane into regions. Then the area of bounded region is

•  200 sq. units^°
•  400 sq. units^°
•  800 sq. units^°
•  500 sq. units^°

Solution

Q3.  The number of values of 'a' for which the lines
2x+y-1=0,
ax+3y-3=0, and
3x+2y-2=0
Are concurrent is

•  0
•  1
•  2
•  Infinite

Solution

Q4. Locus of the image of the point (2, 3) in the line (x-2y+3)+λ(2x-3y+4)=0 is (λ∈R)

•  x²+y²-3x-4y-4=0
•  2x²+3y²+2x+4y-7=0
•  x²+y²-2x-4y+4=0
•  None of these

Solution

Q5.Equations of diagonals of square formed by lines x=0,y=0,x=1 and y=1 are

•  y=x,y+x=1
•   y=x,x+y=2
•  y=x,x+y=2
•   y=2x,y+2x=1

Solution

Q6. A pair of perpendicular straight lines is drawn through the origin forming with the line 2x+3y=6 an isosceles triangle right angled at the origin. The equation to the line pair is

•  5x²-24xy-5y²=0
•  5x²-26xy-5y²=0
• 5x²+24xy-5y²=0
•  5x²+26xy-5y²=0

Solution

Q7.The line PQ whose equation is x-y=2 cuts the x-axis at P and Q is (4, 2). The line PQ is rotated about P through 45° in the anticlockwise direction. The equation of the line PQ in the new position is

•  y=-√2
•  y=2
•  x=2
•  x=-2

Solution

Q8.If the slope of one line represented by a³ x²-2hxy+b³ y²=0 is square of the slope of another line, then

•  h=2ab(a+b)
•  h=ab(a+b)
•  3h=2ab(a+b)
•  2h=ab(a+b)

Solution

Q9.The equations of the sides of a triangle are x+y-5=0,x-y+1=0 and y-1=0. Then the coordinates of the circumcentre are

•  (2, 1)
•  (1, 2)
•  (2,-2)
•  (1,-2)

Solution

Q10. A beam of light is sent along the linex-y=1, which after refraction from the x-axis enters the opposite side by running through 30° towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is

•  (2-√3)x-y=2+√3
•  (2+√3)x-y=2+√3
•  (2-√3)x+y=(2+√3)
• y=(2-√3)(x-1)

Solution