Foot of Perpendicular from a Point on a Line
Trending Questions
Q.
If PM is the perpendicular from P (2, 3) on to the line x+y=3 then the co-ordinates of M are
(2, 1)
(−1, 4)
(4, −1)
(1, 2)
Q. Find the coordinates of the foot of perpendicular from the point ( – 1, 3) to the line 3 x – 4 y – 16 = 0.
Q. find the length of the diameter of the circle which passes through points (-3, 6), (-5, 3), (3, -6)
Q. The perpendicular from the origin to a line meets it at the point (– 2, 9), find the equation of the line.
Q. For non-negative real numbers h1, h2, h3, k1, k2, k3, if the algebraic sum of the perpendiculars drawn from the points (2, k1), (3, k2), (7, k3), (h1, 4), (h2, 5), (h3, −3) on a variable line passing through (2, 1) is zero, then
- h1=1, h2=1, h3=2
- k1=k2=k3=0
- h1=h2=h3=0
- k1=1, k2=2, k3=3
Q. 3. If a straight line is perpendicular to 2x+8y = 10 and meets the x - axis at (5, 0), then it meets the y - axis at ?
Q. Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point:
- (2, −12)
- (1, 34)
- (1, −34)
- (2, 12)
Q.
The coordinates of the image of the origin O with respect to the straight line x+y+1=0 are
(−12, −12)
- (−2, −2)
(1, 1)
(−1, −1)
Q. 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
Q. A straight line passes through a fixed point (h, k). The locus of the foot of perpendicular on it drawn from the origin is
- x2+y2=hx−ky
- x2+y2=kx−hy
- x2+y2−hx−ky=0
- x2+y2−kx−hy=0