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Maths ques ... again

Ques b.ii) and working (sorry i drew another diagram in pencil).
14571181703931837717834.jpg
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You know that the gradient of mAB→=ΔyΔx=−32m_{\overrightarrow{AB}} = \frac{\Delta y}{\Delta x} = - \frac{3}{2}
You know that the line connecting B and C is orthogonal to this line. Therefore mBC→=23m_{\overrightarrow{BC}}= \frac{2}{3}
You can use this to find kk
Reply 2
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B_9710
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It's important to realise that point C will lie on the line that you gave in part bi)
And the y coordinate of C is 7.
Reply 3
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TeeEm
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am I too late?
Original post by TeeEm
am I too late?


i think so just like me again ;(
Reply 5
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kiiten
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Original post by B_9710
It's important to realise that point C will lie on the line that you gave in part bi)
And the y coordinate of C is 7.


Original post by MathsAndChess
You know that the gradient of mAB→=ΔyΔx=−32m_{\overrightarrow{AB}} = \frac{\Delta y}{\Delta x} = - \frac{3}{2}
You know that the line connecting B and C is orthogonal to this line. Therefore mBC→=23m_{\overrightarrow{BC}}= \frac{2}{3}
You can use this to find kk


I subbed in 7 to get k = 5 but Im confused - how do you know c lies on this line?
(edited 8 years ago)
Reply 6
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Zacken
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Original post by kiiten
I subbed in 7 to get k = 5 but Im confused - how do you know c lies on this line?


Right angle = 90 degrees.

Perpendicular = 90 degrees.

Ringing any bells?
Original post by kiiten
Ques b.ii) and working (sorry i drew another diagram in pencil).
14571181703931837717834.jpg
Attachment not found


draw a sketch and you'll simply spot that for the condition that ABC is 90 is must on lie on the line perpendicular to AB
Reply 8
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kiiten
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Original post by Zacken
Right angle = 90 degrees.

Perpendicular = 90 degrees.

Ringing any bells?


Yes. Thanks :smile:

What about finding the distance from a circle to a tangent at a point (outside the circle). Would you just find the distance from centre to the point then minus the radius. - I got confused on the 'tangent' part.

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Reply 9
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Zacken
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Original post by kiiten
Yes. Thanks :smile:

What about finding the distance from a circle to a tangent at a point (outside the circle). Would you just find the distance from centre to the point then minus the radius. - I got confused on the 'tangent' part.

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Sketch it, use the fact that the angle between the radius and the tangent at a point is 90 degrees, then use trigonometry or Pythagoras. :smile:
Reply 10
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kiiten
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Original post by Zacken
Sketch it, use the fact that the angle between the radius and the tangent at a point is 90 degrees, then use trigonometry or Pythagoras. :smile:


Is it like this (finding the hypotenuse)? The length of PC is bigger than the radius and i dont think P lies on the circle.

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1457123321621.jpg18.8KB
Reply 11
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Zacken
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Original post by kiiten
Is it like this (finding the hypotenuse)? The length of PC is bigger than the radius and i dont think P lies on the circle.

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Not quite, more like this:



Except only one half is rquired and none of those lines inside the circle. Just the one line from the centre to the circumference (the radius line) and then the tangent to the point. The angle CAP is a right angle.
Reply 12
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kiiten
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Original post by Zacken
Not quite, more like this:



Except only one half is rquired and none of those lines inside the circle. Just the one line from the centre to the circumference (the radius line) and then the tangent to the point. The angle CAP is a right angle.


Ohh so the tangent could be either AP or BP?

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Reply 13
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Zacken
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Original post by kiiten
Ohh so the tangent could be either AP or BP?

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Yes, your sketch can be either one of those. I just couldn't find a proper picture that showed only one, block half of it our, sorry. :colondollar:
Reply 14
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kiiten
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Original post by Zacken
Yes, your sketch can be either one of those. I just couldn't find a proper picture that showed only one, block half of it our, sorry. :colondollar:


Dw its fine, at least you made the effort. I understood what you were trying to say. Thank you!! :biggrin:

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