Foot of Perpendicular from a Point on a Line
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- (2, 0, 1)
- (−1, 0, 4)
- (4, 0, −1)
- (1, 0, 2)
- √32
- 12√3
- 1√3
- 1√5
The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are x−y+5=0 and x+2y=0 respectively. If the co-ordinates of A are (1, −2), then the equation of BC is
- 23x+14y−40=0
- 14x+23y−40=0
- 23x−14y+40=0
- None of these
The coordinates of the foot of the perpendicular from the point (2, 3) on the line x+y−11=0 are
(-6, 5)
(5, 6)
(-5, 6)
(6, 5)
- 1:3
- 3:1
- 1:9
- 9:1
- x2+y2=c2
- x2+y2=(a2−b2)2
- x2+y2=a2
- x2+y2=b2
- (–4, –3)
- (3, 4)
- (–4, 3)
- (–3, 4)
The equation represents a parabola whose latus rectum is
- (−1, −1)
- (1, 1)
- (1, −1)
- (−1, 1)
A line forms a triangle of area square units with the coordinate axes. Find the equation of the line if the perpendicular drawn from the origin to the line makes an angle of with theaxis
- (75, 258)
- (825, 75)
- (115, 85)
- (5, 87)
Find the coordinates of the foot of the perpendicular from the point (−1, 3) to the line 3x−4y−16=0.
- 2a211 sq. units
- 2a29 sq. units
- a27 sq. units
- 2a27 sq. units
Find the projection of the point (1, 0) on the line joining the points (−1, 2) and (5, 4).
- (-2, -8)
- (2, 20)
- (-1, -4)
- (1, 10)
Find the direction cosines of the line segment joining the two points (1, - 2, 3). and ( 2, 4, 5)