Question

Let P and Q be the points of trisection of the line segment joining the points A(2, – 2) and B(–7, 4) such that P is nearer to A. The coordinates of P and Q are given by​

Answer 1

Answer:

p(-1,0) and Q(-4,2)

Step-by-step explanation:

dividing the line AB in three equal parts

Answer 2

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

Given that,

P and Q be the points of trisection of the line segment joining the points A(2, – 2) and B(–7, 4) such that P is nearer to A.

Case :- 1

So, it means P divides the line segment joining the points A (2, - 2) and B ( - 7, 4) in the ratio 1 : 2.

We know,

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

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

So, on substituting the values, we get

$\rm \: Coordinates\:of\:P$

$\rm \: = \: \bigg(\dfrac{1( - 7) + 2(2)}{1 + 2} ,\dfrac{1(4) + 2( - 2)}{1 + 2} \bigg)$

$\rm \: = \: \bigg(\dfrac{ - 7 + 4}{3} ,\dfrac{0}{3} \bigg)$

$\rm \: = \: ( - 1,0)$

So,

$\rm\implies \:\boxed{\sf{ \:Coordinates\:of\:P = \: ( - 1,0) \: }}$

Now, Case :- 2

Q is the midpoint of the line segment joining the points P ( - 1, 0) and B ( - 7, 4).

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, on substituting the values, we get

$\rm \: Coordinates\:of\:Q$

$\rm \: = \: \bigg(\dfrac{ - 1 + ( - 7)}{2} ,\dfrac{0 + 4}{2} \bigg)$

$\rm \: = \: \bigg(\dfrac{ - 1 - 7}{2} ,\dfrac{4}{2} \bigg)$

$\rm \: = \: \bigg(\dfrac{ -8}{2} ,2 \bigg)$

$\rm \: = \: ( - 4,2) \:$

So,

$\rm\implies \:\boxed{\sf{ \:Coordinates\:of\:Q = \: ( - 4,2) \: }}$

Hence,

$\rm\implies \:\boxed{\sf{ \:Coordinates\:of\:P = \: ( - 1,0) \: }}$

and

$\rm\implies \:\boxed{\sf{ \:Coordinates\:of\:Q = \: ( - 4,2) \: }}$

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Additional Information :-

1. 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

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

2. 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

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

3. 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

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

4. 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

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