Perpendicular Form of a Straight Line
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Q. A line passes through (2, 3) and the portion of it intercepted between co-ordinate axes is divided in the ratio 2:3 at the point (2, 3). Then equation of the line in Normal form is
- xcosπ4+ysinπ4=5√2
- xcosπ3+ysinπ3=3√2
- 9√97x+4√97y=30√97
- 3√13x+2√13y=6√13
Q. The equation of the line perpendicular to the line 2x+3y+5=0 and passing through (1, 1), is
- 3x−2y+1=0
- 3x−2y=2
- 3x−2y=1
- 3x−2y+2=0
Q. Let PQRS is a parallelogram where P=(2, 2), Q=(6, −1), and R=(7, 3). Then equation of the line through S and perpendicular to QR is
- x+4y=27
- 4x−3y+6=0
- x+4y=10
- x+4y=2
Q. Consider two points A(1, a) and B(5, b). If the equation of the line bisecting the line segment AB perpendicularly is x−3y=0 then |ab| is
Q. The perpendcular form of the straight line
y=x+4 is xcosα+ysinα=p then
y=x+4 is xcosα+ysinα=p then
- α=45∘
- α=135∘
- p=2√2
- p=√2
Q. If two lines perpendicular to each other always satisfy the condition cotθ1+cotθ2=2, where θ1 and θ2 are the angles made by the lines with positive direction of X−axis and M1, M2 be the respective slopes, then ∣∣M1+√2M2∣∣ is equal to
(where M1>M2)
(where M1>M2)
Q. The perpendicular distance of a line from origin is 2 units and the perpendicular makes an angle α with X-axis such that sinα=13. The equation of line is .
- 2√2x+y=6
- 2√2x−y=6
- x+2√2y=6
- x−2√2y=6
Q. A square of side length a units lies above the x−axis and has one vertex at the origin. The side passing through the origin makes an angle α(0<α<π/4) with the positive direction of x−axis. The equation of its diagonal not passing through the origin is
- y(cosα+sinα)+x(sinα−cosα)=a
- y(cosα+sinα)+x(sinα+cosα)=a
- y(cosα+sinα)+x(cosα−sinα)=a
- y(cosα−sinα)−x(sinα−cosα)=a
Q. The line perpendicular to y=√3x+2 and passing through (2, 1) is rotated through an angle 60∘ about (2, 1). Area of triangle formed by these lines and x-axis is A units2, then 4A2 is :