Question

Find the midpoint of(3,0)and(1,4)​

Answer 1

Answer:

(2,2)

Step-by-step explanation:

given points, (3,0) & (1,4)

=>  x₁ = 3 ,y₁ = 0 &  x₂ = 1, y₂=4

midpoint of 2 points (x₁y₁) & (x₂,y₂) is given by M ,($\frac{x_{1}+x_{2} }{2} , \frac{y_{1}+y_{2} }{2}$)

=> midpoint is ($\frac{3+1}{2} ,\frac{0+4}{2}$)

                    = ($\frac{4}{2} , \frac{4}{2}$)

                    = (2,2)

Answer 2

We know that :-

Co-ordinates of midpoint of line joining (a,b) and (c,d) is given as :-

$\rm\boxed{\boxed{ \rightarrow \bigg( \large \tt\dfrac{a + c}{2} \: , \: \dfrac{b + d}{2} \bigg)}}$

Let point P (k,m) be midpoint of line joining (3,0) and (1,4).

Therefore , co-ordinates of P(k,m) is given as :-

$\tt \longrightarrow\bigg( k , m\bigg) = \bigg( \dfrac{3 + 1}{2} \: , \: \dfrac{4 + 0}{2} \bigg)$

$\tt \longrightarrow\bigg( k , m\bigg) = \bigg( \dfrac{4}{2} \: , \: \dfrac{4}{2} \bigg)$

$\tt \longrightarrow( k , m) = ( 2 ,2)$

Therefore , midpoint of line joining (3,0) and (1,4) = (2,2)

$\tt \large \red{Additional-Information :}$

$\tt \: Equation \: of \: line \: passing \: through \: points \: (x_1 , y_1) \: and \: (x_2 , y_2) :$

$\tt \longrightarrow y - y_1 = x-x_1\bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg )$

$\tt \rightarrow \: here \: \bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg ) is \: slope \: of \: line$

=> Area of triangle when it's vertices are given :-

$\rightarrow \tt Area \: of \: triangle = \sf{\dfrac{1}{2} \times \left|\begin{array}{c c c} \tt x_{1} & \tt y_{1} & \tt 1 \\ \tt x_{2} & \tt y_{2} & \tt 1 \\ \tt x_{3} & \tt y_{3} & \tt 1\end{array}\right|}$