Find the line perpendicular to [tex]$y=-\frac{1}{2} x-1$[/tex] that includes the point (-2,4).
[tex]$y-[?]=\square(x-\square)$[/tex]
Answer (3)
Well, create a right triangle so that the segment you want to measure is the hypotenuse. The legs should be parallel to the two axes. The triangle could be degenerate (one of the legs has length 0). Find out the length of the legs, then, using Pythagoras' Theorem, find the hypotenuse.
Let's say we have a right triangle with vertices at A(0,0), B(10,0), and C(10,10).
As you can see, the triangle has side lengths of 10 and 10.
Let's now find the distance of the hypotenuse.
AC = 1 0 2 + 1 0 2
As you can see, this provides us the distance of the hypotenuse, or the line segment AC.
If we look at AC closer, we notice that AC is represented by points on the coordinate plane.
Now, try finding the distance between points A and C using the distance formula.
Distance = ( 10 − 0 ) 2 + ( 10 − 0 ) 2
Distance = 1 0 2 + 1 0 2
As this is the same as AC, we can say that we have now derived the distance formula from the Pythagorean theorem. ;
The Distance Formula can be justified using the Pythagorean Theorem by considering the straight-line distance between two points as the hypotenuse of a right triangle formed by the differences in their x and y coordinates. Applying the Pythagorean Theorem leads directly to the Distance Formula, d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 . Thus, the relationship clearly illustrates how distance in a 2D space is calculated through this geometric principle.
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