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# Question

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1. looking at the picture
can some please tell me where does the author get the (-3) x 4 from.

thank you
2. (Original post by bigmansouf)
looking at the picture
can some please tell me where does the author get the (-3) x 4 from.

thank you
So for two lines to be perpendicular to each other, their gradients must times to equal -1.

So for two general lines ax + by + c = 0 and dx + ey + f = 0, the gradients are -a/b and -d/e respectively.

Therefore (-a/b) * (-d/e) = -1, which implies ad = - be or ad + be = 0

In this example a = k+1, b = -3, d = k-2, e = 4

Hence, (k+1)*(k-2) + (-3) *(4) = 0

Hope this makes it clear where all the terms come from
3. (Original post by bigmansouf)
looking at the picture
can some please tell me where does the author get the (-3) x 4 from.

thank you
This is how I think they have got it
and

For the lines to be perpendicular:

Therefore,

Don't know if this helps.
4. (Original post by xylas)
So for two lines to be perpendicular to each other, their gradients must times to equal -1.

So for two general lines ax + by + c = 0 and dx + ey + f = 0, the gradients are -a/b and -d/e respectively.

Therefore (-a/b) * (-d/e) = -1, which implies ad = - be or ad + be = 0

In this example a = k+1, b = -3, d = k-2, e = 4

Hence, (k+1)*(k-2) + (-3) *(4) = 0

Hope this makes it clear where all the terms come from
thank you
5. (Original post by Cryptokyo)
This is how I think they have got it
and

For the lines to be perpendicular:

Therefore,

Don't know if this helps.
thanks you
6. For part two the roots of are

And,

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