Perpendicular Distance of a Point from a Line
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The perpendicular bisector of the line segment joining and has intercept Then, the possible value of is
Prove that the product of the lengths of the perpendiculars drawn from the points (√(a2−b2, 0) and (−√(a2−b2, 0) to the line
xacos θ+ybsin θ=1 is b2
- x2+y2+18x−28y+27=0
- x2+y2+24x−27y+26=0
- x2+y2+48x−12y+11=0
- x2+y2+18x−22y+21=0
A parabola passing through the point (-4, -2) has its vertex at the origin and y-axis as its axis. The latus rectum of
the parabola is
8
10
12
6
- 4y+9=0
- y−2=0
- y−3=0
- 6y−13=0
- x1, x3, x2 are in A.P.
- x1, x3, x2 are in G.P.
- y1, y3, y2 are in A.P.
- y1, y3, y2 are in G.P.
Match the lines given on the left side with their corresponding slopes on the right..
Line passes through the pointsSlope of the linep.) (1, 6) and (-4, 2)1.) 0q.) (5, 9) and (2, 9)2.) -3r.) (-2, -1) and (-3, 2)3.)45s.) (4, 0) and (3, 3)4.)53
p - 1 q - 3 r - 1 s - 2
p - 1 q - 3 r - 2 s - 2
p - 3 q - 1 r - 2 s - 2
p - 3 q - 1 r - 2 s - 4
- x+y+√2=0
- 3x+4y+5=0
- x+y−√2=0
- 3x+4y−5=0
Find the equation for the ellipse that satisfies the given conditions,
Vertices (±5, 0), Foci(±4, 0)
The coordinates of a point at unit distance from the lines 3x - 4y + 1 = 0 and 3x + 6y + 1 = 0 are
(65, −110)
(0, −110)
(−25, −1310)
(−85, 310)
- −2
- −4
- 2
- 1
- 1p2=1a2+1b2
- 1p2=1a2−1b2
- 1p2=a2b2a2+b2
- 1p2=−1a2+1b2
Let the algebraic sum of the perpendicular distance from the points (2, 0), (0, 2), and (1, 1) to a variable straight line be zero. Then the line passes through a fixed point whose coordinate are
(0, 0)
(3, 3)
(1, 1)
(2, 2)
Find the value of p so that three lines 3x + y - 2 = 0, px + 2y - 3 = 0 and 2x - y - 3 = 0 may intersect at one point.
- 43(√3+2)
- 43(2−√3)
- 23(√3+2)
- 23(2−√3)
Find the number of solutions for y=−2x+2 and y = −8x + 8.
- x=−3
- x=−12
- x=−8
- x=−4
- 1
- -2
- 2
- -4
- (3, 1), (−7, 11)
- (3, 1), (−9, 13)
- (2, 2), (−8, 12)
- (−7, 11), (−17, 21)
- 1
- 2
- 3
- 4
(i) (−3, 2) and (1, 4)
(ii)
(iii) (3, −5), and (1, 2)