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# How to attempt finding unknown y coord in a triangle

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1. question b) is the one I do not know how to approach. I know I could probably use pythagoras, but I predict that the calculations will get a little bit messy. There are other ways to complete this on the mark scheme, like "gradient LM * gradient MN", but I don't understand where they're coming from.

any help pls?
2. No Pythagoras involved. The line MN passes through M and is perpendicular to ML. Find the equation of the line passing through both M and N and then you should see how to find p.

To find p, plug x=16 I to the equation of the line MN that you found.
3. if you know the gradient of LM you can find the gradient of MN as it is perpendicular...
4. (Original post by frostyy)

question b) is the one I do not know how to approach. I know I could probably use pythagoras, but I predict that the calculations will get a little bit messy. There are other ways to complete this on the mark scheme, like "gradient LM * gradient MN", but I don't understand where they're coming from.

any help pls?
If (and only if) two lines are perpendicular, the product of their gradients is -1. (You can prove this using vectors).
Thus gradient LM = (-4-2)/(7+1) = -6/8 = -3/4, and gradient MN = (p+4)/(16-7) = (p+4)/9.
Since angle LMN = 90 degrees, LM is perpendicular to MN.
Thus -3/4 * (p+4)/9 = -1 -> 3(p+4)/36 = 1 -> p+4 = 12 -> p=8.
5. (Original post by HapaxOromenon2)
If (and only if) two lines are perpendicular, the product of their gradients is -1. (You can prove this using vectors).
Thus gradient LM = (-4-2)/(7+1) = -6/8 = -3/4, and gradient MN = (p+4)/(16-7) = (p+4)/9.
Since angle LMN = 90 degrees, LM is perpendicular to MN.
Thus -3/4 * (p+4)/9 = -1 -> 3(p+4)/36 = 1 -> p+4 = 12 -> p=8.
Okay, thank you, I get that.

What about part c)? I know the answer is 2 + p + 4, but I'm not sure why?
6. (Original post by frostyy)
Okay, thank you, I get that.

What about part c)? I know the answer is 2 + p + 4, but I'm not sure why?
Does this help at all:

7. (Original post by Zacken)
Does this help at all:

That's how I illustrated it to myself at the start, but I still can't see why you'd add the present 3 y-coords (+ ignore the minus sign at M (7, -4)?).
8. (Original post by frostyy)
That's how I illustrated it to myself at the start, but I still can't see why you'd add the present 3 y-coords (+ ignore the minus sign at M (7, -4)?).
Would it help if you think about it as vectors? (start from M, add the vector MN then add the vector -ML)

NB: Another way of doing this would be to write then use the gradients perpendicular to LM and parallel to MN to get a simultaneous equation in x,y.
9. (Original post by Zacken)
Would it help if you think about it as vectors? (start from M, add the vector MN then add the vector -ML)
Okay, that makes sense, thank you. Woah, I'd never expect to use vectors in C1.

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