Question

Find the coordinates of the point of trisection of the line segment joining the points (3,-1) and (6,8).
Give correct answer then I will mark as brainliest.

Answer 1

Answer:

(15/4,5/4)

Step-by-step explanation:

use the formula (mx2+nx1/m+n,my2+ny1/m+n)

m and n are 1 ,3

hope it helps you.

Answer 2

Question :-

Find the coordinates of the point of trisection of the line segment joining the points (3,-1) and (6,8) .

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 , - 1 )
• F has coordinates ( 6 , 8 )

⇰ Finding coordinates of point A by section formula

EA  : AF  = 1 : 2

so,

$\sf{(p,q)=\left(\frac{(1)(6)+(2)(3)}{(1)+(2)}\:,\:\frac{(1)(8)+(2)(-1)}{(1)+(2)}\right)}\\\\\sf{(p,q)=\left(\frac{6+6}{3}\:,\:\frac{8-2}{3}\right)}\\\\\boxed{\large{\sf{\purple{(p,q)=(4,2)}}}}$

⇰ Finding coordinates of point B by section formula

Now we get ,

( p , q ) = ( 4 , 2 )

AB : BF = 1 : 1

so,

$\sf{(m,n) = \left(\frac{(1)(6) + (1)(4)}{(1)+(1)}\:,\:\frac{(1)(8)+(1)(2)}{(1)+(1)}\right)}\\\\\sf{(m,n)=\left(\frac{6+4}{2}\:,\:\frac{8+2}{2}\right)}\\\\\boxed{\large{\sf{\purple{(m,n)=(5,5)}}}}$

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Hence, points of trisection of line segment joining the points (3,-1) and (6,8) are ( 4 , 2) and ( 5 , 5 ) .