Math, asked by Lovelysekhar, 10 months ago

Find the coordinates of the points trisection of the line segment joining the points.
1.(-3,3)and(3,-3)

Answers

Answered by manikiran18
5

hlo mate

Trisection means a line is dividing into 3 equal parts.

This can be done by finding two points P and Q on the line segment AB. Such that AP=PQ=QB.

Let AP= PQ= QB=x

Then, AP= x & PB= PQ+QB= x+x=2x

AP:PB= x:2x= 1:2

AQ= AP+PQ= x+x=2x & QB= x

AQ:QB= 2x:x= 2:1

Hence, to find points of trisection we find two points P and Q such that P divides AB

in the ratio 1 : 2 and Q divides AB  in the ratio 2:1

solution is in the attatchment

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Answered by SugarCrash
44

Question :-

Find the coordinates of the point of trisection of the line segment joining the points (3-,3) and (3,-3) .

Formula required :-

▶ Section formula

\boxed{\bf{(x,y)=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2}\:,\:\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)}}

where ,P ( x , y ) giving the coordinates of point dividing point A ( x₁ , y₁ ) and point B ( x₂ , y₂ ) in the ratio m₁ : m₂ .

Solution :-

Refer to the figure firstly ...

Where A and B are points of trisection of line EF

( p , q ) are the coordinate of point A

( m , n ) are coordinate of point B

☞ E has coordinates (-3 ,3 )

☞ F has coordinates ( 3 , -3 )

⇰ Finding coordinates of point A by section formula

EA : AF = 1 : 2

so,

\sf{(p,q)=\left(\frac{(1)(3)+(2)(-3)}{(1)+(2)}\:,\:\frac{(1)(-3)+(2)(3)}{(1)+(2)}\right)}\\\\\sf{(p,q)=\left(\frac{3-6}{3}\:,\:\frac{-3+6}{3}\right)}\\\\ \sf{(m,n)=\left(\frac{-3}{3}\:,\:\frac{3}{3}\right)}\\\\\red\bigstar\underline{\boxed{\large{\mathfrak{\purple{(p,q)=(-1,1)}}}}}

⇰ Finding coordinates of point B by section formula

Now we have ,

☞ Coordinates of point A( p , q ) = ( -1 , 1 )

☞ Coordinates of point F = (3 , -3)

AB : BF = 1 : 1

so,

\sf{(m,n)=\left(\frac{(1)(3)+(1)(-1)}{(1)+(1)}\:,\:\frac{(1)(-3)+(1)(1)}{(1)+(1)}\right)}\\\\\sf{(m,n)=\left(\frac{3-1}{2}\:,\:\frac{-3+1}{2}\right)}\\\\\sf{(m,n)=\left(\frac{2}{2}\:,\:\frac{-2}{2}\right)}\\\\\red\bigstar\underline{\boxed{\large{\mathfrak{\purple{(m,n)=(1,-1)}}}}}

_________________________

Hence,

Points of trisection of line segment joining the points(-3, 3) and (3,-3) are ( -1 ,1 ) and ( 1, -1) .

{\fcolorbox{red}{blue}{\orange{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: SugarCrash࿐\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}} 

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