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What is the perpendicular distance from the point 2 and 4 to the straight line equation of y equals 2x plus 10 showing key stages of work?
1 Coordinates: (2, 4) 2 Equation: y = 2x+10 3 Perpendicular equation: y = -0.5+5 4 They intersect at: (-2, 6) 5 Distance is the square root of: (-2, -2)2+(6, -4) = 2*sq rt of 5 = 4.472 to 3 decimal places
What is the perpendicular distance from the point 19 29 that meets the straight line equation 3y equals 9x plus 18 on the Cartesian plane showing work and answer to an appropriate degree of accuracy?
If: 3y = 9x+18 then y = 3x+6 with a slope of 3 Perpendicular slope: -1/3 Perpendicular equation: y-29 = -1/3(x-19) => 3y = -x+106 Both equations intercept at: (8.8, 32.4) Perpendicular distance: square root of (8.8-19)^2+(32.4-29)^2 = 10.75 rounded
What is the perpendicular distance from the point of 7 and 5 that meets the straight line equation of 3x plus 4y equals 0 on the Cartesian plane showing work?
Equation: 3x+4y = 0 => y = -3/4x Perpendicular slope: 4/3 Perpendicular equation: 4x-3y-13 = 0 Equations intersect at: (2.08, -1.56) Distance from (7, 5) to (2.08, -1.56) = 8.2 units using the distance formula
What is the perpendicular distance from the point 4 -2 to the line 2x -y -5 equals 0 showing key stages of work?
Points: (4, -2) Equation: 2x-y-5 = 0 Perpendicular equation: x+2y = 0 Equations intersect at: (2, -1) Perpendicular distance is the square root of: (2-4)2+(-1--2)2 = 5 Distance = square root of 5
What is the perpendicular distance from the point of 2 4 on the Cartesian plane to the straight line equation of y equals 2x plus 10 showing key stages of work?
1 Equation: y = 2x+10 2 Perpendicular equation works out as: 2y = -x+10 3 Point of intersection: (-2, 6) 4 Distance is the square root of: (-2-2)2+(6-4)2 = 2 times sq rt of 5
What is the distance from the point 5 7 that is perpendicular to to the straight line equation of 3x-y plus 2 equals 0 on the Cartesian plane showing key aspects of work?
To find the perpendicular distance from a given point to a given line, find the equation of the line perpendicular to the given line which passes through the given point. Then the distance can be calculated as the distance from the given point to the point of intersection of the two lines, which can be calculated by using Pythagoras on the Cartesian coordinates of the two points. A line in the form y = mx + c has gradient m. If a line has gradient m, the line perpendicular to it has gradient m' such that mm' = -1, ie m' = -1/m (the negative reciprocal of the gradient). A line through a point (x0, y0) with gradient m has equation: y - yo = m(x - x0) Thus the equation of the line through (5, 7) that is perpendicular to 3x - y + 2 = 0 can be found. The intercept of this line with 3x - y + 2 = 0 can be calculated as there are now two simultaneous equations. → The perpendicular distance from (5, 7) to the line 3x - y + 2 = 0 is the distance form (5, 7) to this point of interception, calculated via Pythagoras: distance = √((change_in_x)^2 + (change_in_y)^2) This works out to be √10 ≈ 3.162
What is the perpendicular distance from the point 7 5 to the staight line equation 3x plus 4y -16 equals 0 showing key stages of work?
Equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Both equations intersect at: (4, 1) Perpendicular distance: square root of (7-4)2+(5-1)2 = 5
What is the nature of the distance time graph for uniform motion of an object?
The distance-time graph for uniform motion of an object is a straight line with a constant slope. This indicates that the object is covering equal distances in equal time intervals, showing a constant speed.
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 distance from the point 7 5 that meets the line of 4 1 and 0 4 at right angles on the Cartesian plane showing key aspects of work?
Points: (4, 1) and (0, 4) Slope: -3/4 Equation: 4y = -3x+16 Perpendicular slope: 4/3 Perpendicular equation: 3y = 4x-13 Both equations meet at: (4, 1) from (7, 5) at right angles Perpendicular distance: square root of [(4-7)squared+(1-5)squared)] = 5 units
What does a distance vs time graph look like for an object experiencing constant acceleration?
A distance vs time graph for an object experiencing constant acceleration would be a straight line that curves upward, showing a steady increase in distance over time.
A car's distance-time graph is a straight libe with a steep slope ow would you describe the car's speed?
A straight line with a steep slope on a distance-time graph indicates that the car is traveling at a high and constant speed. The steepness of the slope represents the rate of distance covered over time, showing that the car is moving quickly. Since the line is straight, it implies that the speed does not change over the observed period.
