we all know that for two straight lines to be perpendicular to each other the product of their slopes must be -1 but in the case of x axis the slope is 0 and for y-axis the slope is not defined and obviously their product is not -1 but we still take them to be perpendicular to each other why?
2006-11-23 13:05:22 · 4 answers · asked by Ramanadhan C 2 in Science & Mathematics ➔ Mathematics
That definition of perpendicular is what is taught at school, but there is actually a better one when you start learning about vectors.
A vector (a,b) just means something which starts at (0,0) and points towards the point (a,b).
Then two vectors (a,b) and (c,d) are perpendicular only if ac+bd = 0.
So in the case of two axes, the vectors are (1,0) and (0,1), so ac + bd = 1*0 + 0*1 = 0.
The good thing about this definition is that it even works in 3d - if you have two lines in 3d pointing in directions (a,b,c) and (d,e,f), they are perpendicular only if ad + be + cd = 0. Etc.
2006-11-23 13:29:57 · answer #1 · answered by stephen m 4 · 1⤊ 0⤋
It is really by definition, but we can construct perpendicular lines with a compass and straightedge. We can construct 2 sets of perpendicular lines and give them the same origin but not super-imposed, so there is and angle between the sets. Each set has a linear representation with respect to the other and if m1 and m2 are the slopes of one set relative to the other then we will have m1m2 = -1
2006-11-23 22:09:59 · answer #2 · answered by Jimbo 5 · 0⤊ 0⤋
very good question. bravo.
if the equations are written in the form
lx+my+n = 0 it all becomes clear.
0 x + 1 y +0 =0 for y = 0
1 x + 0 y + 0 = 0 for x = 0
lines l1 x + m1 y + n1 = 0
and l2 x + m2 y + n2 = 0 are perpendicular iff
li l2 + m1 m2 = 0
and this holds for the example above.
2006-11-24 05:04:27 · answer #3 · answered by paladin 1 · 0⤊ 0⤋
Because they are. If one is vertical and one is horizontal then don't you think that they are perpendicular?
2006-11-23 21:27:19 · answer #4 · answered by modulo_function 7 · 0⤊ 0⤋