Coordinate Plane Worksheets Hidden

Coordinate plane worksheets are designed to help students practice graphing and identifying various points on the coordinate plane. Whether you are an elementary school student just starting to learn about coordinates or a middle school student brushing up on your graphing skills, these worksheets provide a fun and engaging way to master this mathematical concept.

  Hidden Picture Coordinate Graphing Worksheets

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  Cartesian Coordinate Plane

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  Christmas Graph Worksheets

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  Jelly Bean Math Graphing for Kindergarten

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  Coordinate Grid Graph Paper Printable

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  Figurative Language Worksheets 4th Grade

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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  Equations Algebra 1 Test and Review

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Find the point on the coordinate plane where the x-coordinate is 5 and the y-coordinate is -2.

The point on the coordinate plane where the x-coordinate is 5 and the y-coordinate is -2 is (5, -2).

Determine the distance between the points (-3, 4) and (1, -2).

To determine the distance between the points (-3, 4) and (1, -2), you can use the distance formula: d = ?((x? - x?)² + (y? - y?)²), where x? = -3, y? = 4, x? = 1, and y? = -2. Plugging in the values, we get d = ?((1 - (-3))² + (-2 - 4)²) = ?(4² + (-6)²) = ?(16 + 36) = ?52. Therefore, the distance between the points is ?52 or approximately 7.21 units.

Identify the quadrant in which the point (-2, -5) lies.

The point (-2, -5) lies in the third quadrant of the Cartesian coordinate system.

Calculate the slope of the line passing through the points (2, 3) and (-4, 1).

To calculate the slope of the line passing through the points (2, 3) and (-4, 1), we use the formula for slope which is (y? - y?) / (x? - x?). Substituting the values from the points, we get (1 - 3) / (-4 - 2) = -2 / -6 = 1/3. Therefore, the slope of the line passing through these points is 1/3.

Locate the midpoint of the line segment connecting the points (6, 2) and (-4, 8).

To find the midpoint of a line segment, we average the x-coordinates and y-coordinates of the two endpoints. The x-coordinate of the midpoint is (6 + (-4)) / 2 = 1, and the y-coordinate is (2 + 8) / 2 = 5. Therefore, the midpoint of the line segment connecting the points (6, 2) and (-4, 8) is (1, 5).

Determine the equation of the line parallel to the x-axis that passes through the point (3, -3).

Since the line is parallel to the x-axis, the equation would be in the form y = constant, where constant represents the y-coordinate of the point the line passes through. Therefore, the equation of the line parallel to the x-axis passing through the point (3, -3) is y = -3.

Find the equation of the line perpendicular to y = 2x + 1 that passes through the point (-1, 4).

The slope of the line y = 2x + 1 is 2. To find the slope of the line perpendicular to it, we take the negative reciprocal of 2, which is -1/2. Using the point-slope form with the given point (-1, 4), we get the equation of the perpendicular line: y - 4 = -1/2(x + 1), which simplifies to y = -1/2x + 3.

Identify the coordinates of the vertex of the parabola described by the equation y = x^2 - 4x + 3.

To identify the coordinates of the vertex of the parabola described by the equation y = x^2 - 4x + 3, we first write it in the form y = a(x-h)^2 + k where h and k are the coordinates of the vertex. By completing the square for x^2 - 4x, we get: y = (x-2)^2 - 1. Thus, the vertex is at (2, -1).

Determine the equation of the circle with center (2, -3) and a radius of 5 units.

The equation of a circle with center (h, k) and radius r is given by (x - h)^2 + (y - k)^2 = r^2. In this case, with center (2, -3) and a radius of 5 units, the equation of the circle is (x - 2)^2 + (y + 3)^2 = 25.

Find the distance between the point (0, 0) and the line with the equation 3x - 2y = 6.

To find the distance between a point and a line, we can use the formula d = |Ax1 + By1 + C| / ?(A^2 + B^2), where (x1, y1) is the point and the line is Ax + By + C = 0. Plugging in the values, the line equation can be rewritten as 3x - 2y - 6 = 0. Given the point (0, 0), plugging this into the formula yields d = |3(0) - 2(0) - 6| / ?(3^2 + (-2)^2) = 6 / ?13. Therefore, the distance between the point (0, 0) and the line 3x - 2y = 6 is 6/?13 units.