Question

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

Answer 1

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,

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

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Hence,

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

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