Question

8 . If P is the mid-point of the line joining A ( 1, 4 ) and B (3, 6 ) then the co-ordinates of Pis (A) (4, 10) (B) (2, 10) (C) (2, 5) 5 (D) (4, 5)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Coordinates of A (1, 4)

Coordinates of B (3, 6)

and

P is the mid-point of the line joining A ( 1, 4 ) and B (3, 6 ).

Let assume that Coordinates of P be (x, y).

We know, Mid-point formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the mid-point of AB, then the coordinates of C is given by

$\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered}$

So, here

$\rm \:x_1 = \:1$

$\rm \:x_2 = \:3$

$\rm \:y_1 = \:3$

$\rm \:y_2 = \:6$

So, on substituting the values, we get

$\rm \: (x,y) = \bigg(\dfrac{1+3}{2}, \dfrac{4+6}{2}\bigg)$

$\rm \: (x,y) = \bigg(\dfrac{4}{2}, \dfrac{10}{2}\bigg)$

$\rm\implies \:\rm \: (x,y) = (2, \: 5)$

Hence,

$\rm\implies \:\rm \: Coordinates\:of\:P = (2, \: 5)$

So, option (C) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

Distance Formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane, then distance between A and B is given by

$\begin{gathered}\boxed{\tt{ AB \: = \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2} }}} \\ \end{gathered}$

Section formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

$\begin{gathered} \boxed{\tt{ (x, y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

Centroid of a triangle

Centroid of a triangle is defined as the point at which the medians of the triangle meet and is represented by the symbol G.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle and G(x, y) be the centroid of the triangle, then the coordinates of G is given by

$\begin{gathered}\boxed{\tt{ (x, y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}} \\ \end{gathered}$

Area of a triangle

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle, then the area of triangle is given by

$\begin{gathered}\boxed{\tt{ Area =\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|}} \\ \end{gathered}$

Condition for 3 points to be Collinear

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the coordinates in cartesian plane, then points A, B and C are collinear, then

$\begin{gathered}\boxed{\tt{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0}} \\ \end{gathered}$

Answer 2

Answer:

Given :-

• If P is the mid-point of the line joining A(1 , 4) and B(3 , 6).

To Find :-

• What is the co-ordinates of P.

Formula Used :-

$\clubsuit$ Mid-Point Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{M =\: \bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg]}}}\: \: \: \bigstar$

where,

• M = Mid-Point
• (x₁ , y₁) = Co-ordinates of the first point
• (x₂ , y₂) = Co-ordinates of the second point

Solution :-

Given Points :

$\dashrightarrow \bf A =\: (1 , 4)$

$\dashrightarrow \bf B =\: (3 , 6)$

where,

• x₁ = 1
• y₁ = 4
• x₂ = 3
• y₂ = 6

According to the question by using the formula we get,

$\implies \sf\bold{\pink{(x , y) =\: \bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg]}}$

$\implies \sf (x , y) =\: \bigg[\dfrac{1 + 3}{2} , \dfrac{4 + 6}{2}\bigg]$

$\implies \sf (x , y) = \bigg[\dfrac{\cancel{4}}{\cancel{2}} , \dfrac{\cancel{10}}{\cancel{2}}\bigg]$

$\implies \sf (x , y) =\: \bigg[\dfrac{2}{1} , \dfrac{5}{1}\bigg]$

$\implies \sf\bold{\red{(x , y) =\: 2 , 5}}$

$\therefore$ The co-ordinates of P is (2 , 5) .

Hence, the correct options is option no (C) (2 , 5) .