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The only way you can say that is from the general rule that perpendicular lines have
negative reciprocal slopes. You certainly can't demonstrate it from the slopes of the
axes themselves, because the slope of the x-axis is zero, and the slope of the y-axis
is either infinite or else undefined, whichever term bothers you less.
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Why the slope of x-axis and y-axis is not equal to -1 whereas slope of two perpendicular line is -1?
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Why is product of slopes of x axis and y axis not equal to one?
We know that the slope of a line is (Changes in y)/(Changes in x). Does the y-axes has changes in y? No. This means that y-axis does not have a slope. The same thing is for x-axis.
How do you write an equation if the slope is undefined?
Undefined slopes belong to lines that are vertical. These lines do not cross the y-axis, but do cross the x-axis. Therefore, the equation for these lines are always: x = # (where # is the value at which the line is crossing the x-axis).
Why is the product of the slopes of perpendicular lines always negative?
Yes the product will be negative, in fact the product will equal negative one (-1). Think about this. Suppose you have a line y = mx + b, and you want to rotate it 90° counterclockwise. This new line would be like in a coordinate system (x',y') [the x' and y' are called x-prime and y-prime, and differentiate between the original and new coordinate system], where the x' axis runs along the y axis in the positive y direction, and the y' axis runs along the x axis in the negative x direction. So the new line y' = mx' + b with x' = y and y' = -x, is: -x = my + b. Solving for y in the new equation gives y = (-1/m)x - (b/m). So the new slope (-1/m) times the original slope (m) equals (-m/m) = -1, as long as the original slope was not zero.
Which comes first x axis or the y axis?
The x-axis comes first. because x comes before y.
What is the product of the slopes when the x axis and y axis are perpendicular?
The slope of the x-axis is 0 and the y-axis does not have a slope. For all pairs of perpendicular lines, other than those parallel to the axes, the product of their slopes is -1.
Why is the slopes of product x and y axis is -1?
It is not, because the slope of the y-axis is not defined.
What is the product of slope of x and y axes?
The slope of the x-axis is zero.The slope of the y-axis is "undefined" or "infinity". Whichever term you use, it's nota number that can participate in ordinary arithmetic operations. So the product ofthe slopes can't be calculated.For any other two perpendicular lines, the product of their slopes is -1 .
Why the slope of x-axis and y-axis is not equal to -1 whereas slope of two perpendicular line is -1?
0
Why is product of slopes of x axis and y axis not equal to one?
We know that the slope of a line is (Changes in y)/(Changes in x). Does the y-axes has changes in y? No. This means that y-axis does not have a slope. The same thing is for x-axis.
As slopes of x and y axes are 0 and undefined respectively why the product of slopes of x and y axes is -1?
1
What are the slopes of perpendicular lines called?
Slopes of line perpendicular to the x-axis are undefined.
How do you write an equation if the slope is undefined?
Undefined slopes belong to lines that are vertical. These lines do not cross the y-axis, but do cross the x-axis. Therefore, the equation for these lines are always: x = # (where # is the value at which the line is crossing the x-axis).
Is torque independent of location of axis?
No, the axis must be specified: torque = (distance from the axis) X (force). (X is the vector cross-product in this case - meaning the angle also matters.)No, the axis must be specified: torque = (distance from the axis) X (force). (X is the vector cross-product in this case - meaning the angle also matters.)No, the axis must be specified: torque = (distance from the axis) X (force). (X is the vector cross-product in this case - meaning the angle also matters.)No, the axis must be specified: torque = (distance from the axis) X (force). (X is the vector cross-product in this case - meaning the angle also matters.)
Why is the product of the slopes of perpendicular lines always negative?
Yes the product will be negative, in fact the product will equal negative one (-1). Think about this. Suppose you have a line y = mx + b, and you want to rotate it 90° counterclockwise. This new line would be like in a coordinate system (x',y') [the x' and y' are called x-prime and y-prime, and differentiate between the original and new coordinate system], where the x' axis runs along the y axis in the positive y direction, and the y' axis runs along the x axis in the negative x direction. So the new line y' = mx' + b with x' = y and y' = -x, is: -x = my + b. Solving for y in the new equation gives y = (-1/m)x - (b/m). So the new slope (-1/m) times the original slope (m) equals (-m/m) = -1, as long as the original slope was not zero.
Given a line with a slope of -1 over 5 what is the slope of any line that lies perpendicular to it?
If they are perpendicular, the product of their slopes should be -1 -1*X = -1 X = 1 The other line has slope 1
Y equals 5-2x and 2y plus x equals 1 are the lines perpendicular?
Solve both equations for y in terms of x: y = (-2)x+5 y = (-1/2)x+1/2 Multiply the slopes together: (-2) X (-1/2) = 1 In order for the lines to be perpendicular, the product of the slopes would have had to equal -1, but it equals 1, so they're not perpendicular.
