IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: xxemoxx on August 28, 2009, 06:29:25 pm
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Show that the triangle formed by the points (-2,5) , (1,3) , (5,9) is right angled..
Please help
thanks its urgent
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Show that the triangle formed by the points (-2,5) , (1,3) , (5,9) is right angled..
Please help
thanks its urgent
Call the points A,B,C then AB=B-A=(3,-2) and AC=C-A=(7,4) and BC=C-B=(4,6)
dot product AB with BC=3*4+-2*6=0 there AB is at right angles to BC with a right angle at B
=
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thankyou! I slved that Q sometime back
can u help me with this one..
A(7,2) C(1,4) are two vertices of a square ABCD
(a)find the equation of the diagonol BD
(b) Find the coordinates of B and of D
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thanks
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sin(2@+60)=0.3584
find the least 2 +ve values of @...
Pls someone help me.......in solving d sum
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sin(2@+60)=0.3584
find the least 2 +ve values of @...
Pls someone help me.......in solving d sum
this one's a litlle tricky but you can't be fooled by these cie crappy ppl
here u go
sin(2 theta + 60)=0.3584
so
2 theta+60=sin-10.3584
2 theta+60=21.0019...
now comes the part....u know ull get negative answer if u use this value of the inverse
so u take 180-21.0019 and theta + 360 cuz these also have value of inverse is 0.3584 (translation and periodic property)
i hope u got this part
so then,
2 theta+60=158.998...
2 theta= 158.998...-60
theta=49.5
and then
2 theta+60=381.001967...
2 theta= 381.001967-60
theta= 160.5
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thankyou! I slved that Q sometime back
can u help me with this one..
A(7,2) C(1,4) are two vertices of a square ABCD
(a)find the equation of the diagonol BD
(b) Find the coordinates of B and of D
gradient of AC=(4-2)/(1-7)=-1/3
so gradient of perpendicular line=-1/(-1/3)=3
Also the mid point of AC is a point on BD. Mid point of AC=((7+1)/2,(2+4)/2)=(4,3)
Equation of BD is y=3x+c sub (4,3) in to find c
3=3*4+c so c=-9 and y=3x-9 is equation of BD.
the vector from A to C=(-6,2) so the vector from the midpoint is (-3,1) hence since BD is perpendicular the vector from the centre to eith B or D is (1,3) hence the coordinates are (4,3)+(1,3)=(5,6) or (4,3)-(1,3)=(3,0)
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gradient of AC=(4-2)/(1-7)=-1/3
so gradient of perpendicular line=-1/(-1/3)=3
Also the mid point of AC is a point on BD. Mid point of AC=((7+1)/2,(2+4)/2)=(4,3)
Equation of BD is y=3x+c sub (4,3) in to find c
3=3*4+c so c=-9 and y=3x-9 is equation of BD.
the vector from A to C=(-6,2) so the vector from the midpoint is (-3,1) hence since BD is perpendicular the vector from the centre to eith B or D is (1,3) hence the coordinates are (4,3)+(1,3)=(5,6) or (4,3)-(1,3)=(3,0)
sir, wat is this concept named as? i mean, wat is the rule u applied in ur last sentence called?
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Can you quote the part you arequerying
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the vector from A to C=(-6,2) so the vector from the midpoint is (-3,1) hence since BD is perpendicular the vector from the centre to eith B or D is (1,3) hence the coordinates are (4,3)+(1,3)=(5,6) or (4,3)-(1,3)=(3,0)
the above, wat is the concept called? the concept which states abt the sign switching and vectors..just wanted to know so i can look it up and learn it.! plz help
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A vector which goes in the opposite direction has the opposite sign. There is not really a name to it.
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Easy find all the lengths of sides and do
c^2 = a^2 + b^2
taking c as the longest side