As level Pure mathematics Question

[xxemoxx]

Show that the triangle formed by the points (-2,5) , (1,3) , (5,9) is right angled..

Please help
thanks its urgent

[astarmathsandphysics]

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
=

[xxemoxx]

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

[astarmathsandphysics]

thanks

[spiderman]

sin(2@+60)=0.3584
find the least 2 +ve values of @...



Pls someone help me.......in solving d sum

[nid404]

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

[astarmathsandphysics]

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)

[Ghost Of Highbury]

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?

[astarmathsandphysics]

Can you quote the part you arequerying

[Ghost Of Highbury]

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

[astarmathsandphysics]

A vector which goes in the opposite direction has the opposite sign. There is not really a name to it.

[joel]

Easy find all the lengths of sides and do
 
c^2 = a^2 + b^2

taking c as the longest side