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What is the equation of the line segment whose end points are at 2 3 and 11 13 on the Cartesian plane showing work?
Points: (2, 3) and (11, 13) Slope: 10/9 Equation: 9y = 10x+7
What is the equation of a circle whose diameter end points are at 8 7 and 2 -3 on the Cartesian plane?
End points: (8, 7) and (2, -3) Midpoint: (5, 2) which is the center of the circle Radius: square root of 34 Equation of the circle: (x-5)^2 +(y-2)^2 = 34
What is the Cartesian equation of a circle whose end points of its diameter are at 10 -4 and 2 2?
End points: (10, -4) and (2, 2) Midpoint: (6, -1) which is the centre of the circle Distance from (6, -1) to any of its end points = 5 which is the radius Therefore the Cartesian equation is: (x-6)^2 +(y+1)^2 = 25
What is the perpendicular bisector equation of the line segment whose end points are at -2 4 and -4 8 on the Cartesian plane?
End points: (-2, 4) and (-4, 8) Midpoint: (-3, 6) Slope: -2 Perpendicular slope: 1/2 Perpendicular bisector equation: y -6 = 1/2(x--3) => y = 0.5x+7.5
How are points plotted on the cartesian plane?
The Cartesian plane has a horizontal number line which is the x axis and a vertical number line which is the y axis both axes intercept each other at right angles at the pount of origin whose coordinates are (0, 0) Points are plotted on the plane having numerical values for (x, y) and that x is plotted first followed by y
What is the equation and its perpendicular bisector equation of the line whose end points are at -2 3 and 1 -1 on the Cartesian plane?
Points: (-2, 3) and (1, -1) Midpoint: (-0.5, 1) Slope: -4/3 Perpendicular slope: 4/3 Equation: 3y = -4x+1 Perpendicular bisector equation: 4y = 3x+5.5
Which point is at (8 -3)?
11
What is the perpendicular bisector equation of the line whose end points are at s 2s and 3s 8s on the Cartesian plane?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular equation: y -5s = -1/3(x -2s) => 3y = -x +17s Perpendicular bisector equation in its general form: x +3y -17s = 0
What is the equation of a circle whose diameter end points are at 10 -4 and 2 2 on the Cartesian plane?
End points: (10, -4) and (2, 2) Midpoint: (6, -1) Distance from (6, -1) to (10, -4) = 5 Distance from (6, -1) to (2, 2) = 5 Equation of the circle: (x-6)^2 +(y+1)^2 = 25
Is A (1 2) (2 2) (3 2) (4 2) a function or relation in algebra 2?
It is a straight line when plotted on the Cartesian plane whose equation is y=2
What is the center and radius of a circle whose circumference passes through the points 5 0 and 3 4 and -5 0 on the Cartesian plane?
It has centre (0, 0) and radius 5.
Where are all points whose x coordinates equal their y coordinates?
All points whose x-coordinates equal their y-coordinates lie on the line described by the equation (y = x). This line passes through the origin (0, 0) and extends diagonally through the first and third quadrants of the Cartesian plane. Every point on this line has coordinates of the form ((a, a)), where (a) is any real number.
