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What is the point where the axis meets the y axis?
the origin and it has the coordinates of (0,0)
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 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 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 7 3 and -6 1 at its midpoint 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-4 = -13x+6.5 => 2y = -13x+10.5
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 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 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 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
Can you make 2 squares and 4 right-angled triangles using only 8 straight lines?
Draw one square. Join the midpoint of each side to the midpoint of the adjacent sides. The resulting figure meets the requirements.
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
