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Flapdragon
 2 years ago
Best ResponseYou've already chosen the best response.0I know the distance/slope/midpoint formulas, but I'm still kind of confused. :c

anhhuyalex
 2 years ago
Best ResponseYou've already chosen the best response.0find the slope of WY and XZ. If they are perpendicular, they should have slopes that have a product of 1. Slope formula is rise/run.

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.2find slope of the line wy and zx, let it be m1 and m2 let two points be x2,y2 and x1,y1 slope m=(y2y1)/(x2x1) slope of line wy=(4b0)/(aa)=infinity slope of line zx=(bb)/(02a)=0 now find tan inverse of each slope that will give us the angle the line makes with the x axis tan inverse of infinity is 90 degrees tan inverse of 0 is 0 degrees so wy is perpendicular to x axis and zx is parallel to x axis so wy is perpendicular to zx hence the diagonals are perpendicular to each other

anhhuyalex
 2 years ago
Best ResponseYou've already chosen the best response.0@Ash: Precisely. Although a very complicated proof, I can see. :P

Flapdragon
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you so much! That helps a lot. :D

mathmate
 2 years ago
Best ResponseYou've already chosen the best response.0To show in a general way that two lines intersect at 90 degrees can be done by expressing each line in its general form: Ax+By+C=0 Say we have two lines, L1 : 5x+2y5=0, and L2 : 4x + 10y +7=0 If the products of corresponding coefficients of x and y add up to zero, then the two lines intersect at 90 degrees. In the cited example, we find 5(4)+2(10)=0, so we can safely conclude that L1 is perpendicular (orthogonal) to L2. This result comes from linear algebra or vector products. The best part is that the test still works with vertical and horizontal lines: L3 : x  a = 0 L4 : + y  b = 0 Then since 1(0)+0(1)=0, L3 is perpendicular to L4.
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