Question

chalenge for brainly stars:

show that $P(3,m-5)$ is a point of trisection of the line segment joining the pointsA(4,-2) and B(1,4).hence,find the value of m

Answer 1

Your Question:-

Show that $P(3,m-5)$ is a point of trisection of the line segment joining the pointsA(4,-2) and B(1,4).hence,find the value of m

To Find:-

The Value of m

Solution:-

Using Formula:

=>p(x,y)=(mx²+nx1/m+n , my²+ny1/m+n)

=>p(3m-5)=(1×1+2×4/1+2 , 1×4+2×-2/1+2)

=>(1+8/3,4-4/3)

=>(9/3 , 0/3)

=>Corresponding coordinates are equal:

=>m-5=0

=>m=5

Answer 2

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$\huge\mathfrak\pink{Solution :-}$

$\blue\bigstar\underline{\bf{ By \: using \: Section \: formula }}$

$\: \\ \implies \displaystyle \sf{x = \frac{mx2 + nx1}{m + n} \: and \:y = \frac{my2 + ny1}{m + n} }$

$\\ \implies\displaystyle\sf{3=\dfrac{4m+n}{m+n}}$

$\\ \implies\displaystyle\sf{3m+3n=4m+n}$

$\\ \implies\displaystyle\sf{3m+3n-4m-n=0}$

$\\ \implies\displaystyle\sf{-m+2n=0}$

$\\ \implies\displaystyle\sf{m=2n \: n=m/2}$

$\mapsto\sf{y=\dfrac{my_2+ny_1}{m+n}}$

$\mapsto\sf{m-5=\dfrac{-2m+4n}{m+n}}$

$\mapsto\sf{m(m+n)-5(m+n)=-2m+4n}$

$\mapsto\sf{m^2+mn-5m-5n+2m-4n=0}$

$\mapsto\sf{m^2+mn-3m-9n=0}$

$\mapsto\sf{m^2+\dfrac{m(m)}{2}-9m-\dfrac{9m}{2}=0}$

$\mapsto\sf{m^2+\dfrac{m^2}{2}-9m-\dfrac{9m}{2}=0}$

$\mapsto\sf{2m^2+m^2-18m-9m=0}$

$\mapsto\sf{3m^2-27m=0}$

$\mapsto\sf{m(3m)=27m}$

$\mapsto\sf{m=\dfrac{27m}{3m}=\cancel\dfrac{27}{3}=9}$

$\bigstar\underline{\boxed{\sf{the \: value \: of m \: is 9}}}$