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What is the perpendicular equation that meets the line of 7 3 and -6 1 at its midpoint showing work?
Wiki User
∙ 8y ago
Points: (7, 3) and (-6, 1)
Midpoint: (0.5, 2)
Slope: 2/13
Perpendicular slope: -13/2
Perpendicular equation: y-2 = -13/2(x-0.5) => 2y-4 = -13x+6.5 => 2y = -13x+10.5
Wiki User
∙ 8y ago
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What is the equation and its distance from origin that meets the line joining the points 13 17 and 19 23 at its midpoint on the Cartesian plane showing work?
Points: (13, 17) and (19, 23) Midpoint: (16, 20) Slope of required equation: 5/4 Its equation: 4y = 5x or as y = 1.25x Its distance from (0, 0) to (16, 20) = 4 times sq rt 41
What happens where a radius meets the tangent?
The radius and the tangent are perpendicular at the point on the circle where they meet.
What is perpindicular?
A line is perpendicular to another if it meets or crosses it at right angles (90°). Perpendicular means "at right angles". A line meeting another at a right angle, or 90° is said to be perpendicular to it. In the figure above, the line AB is perpendicular to the line DF. If they met at some other angle we would say that AB meets DF 'obliquely'. Move the point A around and create both situations. Move the mouse carefully to get AB exactly perpendicular to DF....
What does y intercept mean?
anywhere a line, on a graph, meets with the y axis
What does x-axis mean?
It's the horizontal line on the Cartesian plane that meets the y-axis at the origin at 90 degrees.
What is the perpendicular bisector equation that meets the line segment of -2 2 and 6 4 at its midpoint showing work?
Points: (-2, 2) and (6, 4) Midpoint: (2, 3) Slope: 1/4 Perpendicular slope: -4 Perpendicular bisector equation: y-3 = -4(x-2) => y = -4x+11
What is the perpendicular bisector equation that meets the line 13 19 and 23 17 at midpoint on the Cartesian plane showing all aspects of work with answer?
Points: (13, 19) and (23, 17) Midpoint: (18, 18) Slope: -1/5 Perpendicular slope: 5 Perpendicular equation: y-18 = 5(x-18) => y = 5x-72
What is the perpendicular equation that meets the line of 2 3 and 5 7 at its midpoint showing key aspects of work?
Here are the key steps:* Find the midpoint of the given line. * Find the slope of the given line. * Divide -1 (minus one) by this slope, to get the slope of the perpendicular line. * Write an equation for a line that goes through the given point, and that has the given slope.
What is the perpendicular bisector equation that meets the line segment of 7 3 and -6 1 showing work in addition to the answer?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular bisector equation: y-2 = -13/2(x-0.5) => 2y = -13x+10.5
What is the perpendicular equation in its general form that meets the line whose coordinates are 2 5 and 11 17 at its midpoint on the Cartesian plane showing all work?
Points: (2, 5) and (11, 17) Midpoint: (6.5, 11) Slope: 4/3 Perpendicular slope: -3/4 Perpendicular equation: y-11 = -3/4(x-6.5) => 4y = -3x+63.5 In its general form: 3x+4y-63.5 = 0
What is the perpendicular bisector equation in its general form that meets the line containing the points 7 3 and -6 1 showing work?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular equation: y-2 = -13/2(x-0.5) => 2y = -13x+10.5 Perpendicular bisector equation in its general form: 26x+4y-21 = 0
What is the equation and its perpendicular distance from the point -3 -5 whose line meets the line of 0 5 to 3 0 at its midpoint on the Cartesian plane showing work?
Points: (0, 5) and (3, 0) Midpoint: (1.5, 2.5) Slope: -5/3 Perpendicular slope: 3/5 Perpendicular equation: y--5 = 3/5(x--3) => 5y = 3x-16 Distance is the square root of (1.5--3)^2+(2.5--5)^2 = 8.746 to three decimal places
What is the perpendicular bisector equation that meets the line of 7 3 and -6 1 on the Cartesian plane?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular equation: y-2 = -13/2(x-0.5 => 2y = -13x+10.5
What is the perpendicular bisector equation that meets the points of 7 3 and -6 1 on the Cartesian plane?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular equation: y-2 =-13/2(x-0.5) => 2y-4 = -13x+6.5 => 2y = -13x+10.5 Therefore the perpendicular bisector equation is: 2y = -13x+10.5
What is the perpendicular bisector equation that meets the line segment of -1 3 and -2 -5?
Midpoint: (-3/2, -1) Gradient or slope: 8 Perpendicular slope: -1/8 Equation: y- -1 = -1/8(x- -3/2) y = -1/8x -3/16 -1 y = -1/8x -19/16 The perpendicular equation can be expressed in the form of: 2x+16y+19 = 0
What is the perpendicular bisector equation that meets the line whose end points are at -2 plus 5 and -8 -3?
Points: (-2, 5) and (-8, -3) Midpoint: (-5, 1) Slope: 4/3 Perpendicular slope: -3/4 Perpendicular bisector equation: y-1 = -3/4(x--5) => 4y-4 = -3x-15 => 4y = -3x-11
What is the equation and its distance from origin that meets the line joining the points 13 17 and 19 23 at its midpoint on the Cartesian plane showing work?
Points: (13, 17) and (19, 23) Midpoint: (16, 20) Slope of required equation: 5/4 Its equation: 4y = 5x or as y = 1.25x Its distance from (0, 0) to (16, 20) = 4 times sq rt 41
