Perpendicular Distance of a Point from a Line
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Given then for all real values of
Find a relation between and such that the point is equidistant from the point and .
Distance between the lines 5x+3y−7=0 and 15x+9y+14=0 is
13 √34
353 √34
352√34
35√34
If for some , the lines and are coplanar, then the line passes through the point
- √3710
- 37√10
- 3710
- √3710
The values of , so that the line touches are
- less than 2
- greater than 2 but less than 3
- greater than 4
- greater than 3 but less than 4
- 5:−1:2
- −5:1:2
- 5:1:−2
- 5:1:2
- 1
- 2
- 3
- 4
If the length of the perpendicular drawn from the origin to the line whose intercepts on the axes are and be , then
If the point P on the curve, is farthest from the point Q, then PQ2 is equal to:
- 6
- 7
- 2√13
- 4√3
- 4a2
- 2a2
- a2
- a22
The equation of the perpendicular bisector of the line segment joining and is
Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to the line x - 2y = 1.
The values of , for which the equation represents a pair of straight lines, is
- 2x+y+3=0
- x−y+6=0
- x+y+6=0
- x−2y+8=0
The three lines are concurrent if
Let be a tangent to the parabola and be a tangent to the parabola such that intersect at right angles. Then meet on the straight line:
The value of , for which the equation represents a pair of straight lines, are
- 25
- 24
- 22
- 20
- 2
- 1
- -2
- -4
- 114√2
- 74√2
- 78
- 2
A point equidistant from the lines , & is
- y2−4ax=(x+a)2cot2α
- y2−4ax=(x−a)2tan2α
- y2−4ax=(x+a)2tan2α
- y2−4ax=(x−a)2cot2α
- x2+y2+24x−27y+26=0
- x2+y2+48x−12y+11=0
- x2+y2+18x−22y+21=0
- x2+y2+18x−28y+27=0
If is the point on , that is closest to then find
The normal at point of the curve is parallel to the line
- m1⋅m2=−1
- m1⋅m2=1
- m1+m2=0
- m1−m2=0