perpendicular distance from the point (x1,y1) to the line ax+by+c=0
is d=(ax1+by1+c)/sqrt(a^2+b^2)
given point (2,4) line equation 2x-y+10=0
d= (2(2)-4+10)/sqrt(2^2+1^2)
d= 10/sqrt(5)
d= 4.472
Straight line equation: y = 2x+10
Perpendicular equation: y = -1/2x+5
Point of intersection: (-2, 6)
Distance: the square root of [(2--2)2+(4-6)2] = 4.472 to 3 decimal places
The lines are perpendicular, and intersect at the point (1.35, 3.55) .
If the graph is a function, no line perpendicular to the X-axis can intersect the graph at more than one point.
(4,1)
(2, 11)
3
It the point is on the line the distance is 0. If the point is not on the line, then it is possible to draw a unique line from the point to the line which is perpendicular to the line. The distance from the point to the line is the distance along this perpendicular to the line.
Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5
the length of a perpendicular segment from the point to the line
Its perpendicular distance.
If you mean the perpendicular distance from the coordinate of (7, 5) to the straight line 3x+4y-16 = 0 then it works out as 5 units.
The length of a line segment that starts at the point and is perpendicular to the original line.
No it is measured from the edge
The perpendicular distance from (2, 4) to the equation works out as the square root of 20 or 2 times the square root of 5
No. It changes by double the (perpendicular) distance from the point to the line.
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point of intersection: (4, 1) Distance: (7-4)2+(5-1)2 = 25 and the square root of this is the perpendicular distance which is 5 units of measurement
From the given information the perpendicular line will form an equation of 2y = -x and both simultaneous line equations will intersect each other at (2,-1) and so distance from (4, -2) to (2, -1) is the square root of 5 by using the distance formula.
perpendicular