the two lines x=ay+b, z= cy+d and x=a'y+b', z=c'y+d' are perpendicular to each other, if: (a) aa'+cc'=1 (b) (a/a') + (c/c') =-1 (c) (a/a')+(c/c')=1 (d) aa'+cc'=-1

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the two lines x=ay+b, z= cy+d and x=a'y+b', z=c'y+d' are perpendicular to each other, if: (a) aa'+cc'=1 (b) (a/a') + (c/c') =-1 (c) (a/a')+(c/c')=1 (d) aa'+cc'=-1

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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the answer would be d since we can rewrite the first two lines as y=(x-b)/a y=(z-d)/c therefore ((x-b)/a)= ((z-d)/c) which gives us similar we write the second two lines is that for giving us and we know if they are perpendicular the dot product gives us 0 thus xx'+1+zz'=0
ok how did you get ??
i mixed up my signs halfway through it should be and and aa'+1+cc'=0

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ohk that works.. thanx bro!

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