Question

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If A and B are ( -2,-2) and ( 2,-4) respectively, find the coordinates of P such that
AP = 3/7AB and P liës on the line segment AB.

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Answer 1

EXPLANATION.

A and B are (-2,-2) and (2,-4).

⇒ AP = 3/7AB.

P lies on the line segment AB.

As we know that,

⇒ AP = 3/7AB.

⇒ AB = AP + PB.

Put the value of AB in equation, we get.

⇒ AP = 3/7(AP + PB).

⇒ 7AP = 3(AP + PB).

⇒ 7AP = 3AP + 3PB.

⇒ 7AP - 3AP = 3PB.

⇒ 4AP = 3PB.

⇒ AP/PB = 3/4.

As we know that,

Formula of :

Section formula,

⇒ x = Mx₂ + Nx₁/M + N.

⇒ y = My₂ + Ny₁/M + N.

Let,

⇒ M = 3  and N = 4.

⇒ x₁ = -2  and  y₁ = -2.

⇒ x₂ = 2  and  y₂ = -4.

Put the value in equation, we get.

⇒ x = 3(2) + 4(-2)/3 + 4.

⇒ x = 6 - 8/7.

⇒ x = -2/7.

⇒ y = 3(-4) + 4(-2)/3 + 4.

⇒ y = -12 - 8/7.

⇒ y = -20/7.

Their Co-ordinates = (-2/7, -20/7).

Answer 2

$\pmb{\sf{Solution:}}$

Given that there are two points:

• A = (-2,-2)

• B = (2,4)

Also, AP = $\sf{\dfrac{3}{7}}$AB

We need to find the coordinate of P.

See the given attachment to get clear with the concept.

As AP = $\sf{\dfrac{3}{7}}$AB

$\implies\sf{BP = AB-AP}$

Substitute AP = $\sf{\dfrac{3}{7}}$AB in $\implies\sf{BP = AB-AP}$

$\implies\sf{BP = AB- \dfrac{3}{7}AB}$

$\implies\sf{BP = \dfrac{7AB-3AB}{7}}$

$\implies\sf{BP = \dfrac{4AB}{7}}$

On dividing AP and BP, we get:

$\implies$ (3/7AB)/(4/7AB)

7 and AB gets cancelled here:

= 3/4

$\implies$AP:BP = 3:4

Hence, we found the ratio!

According to section formula, in x coordinate:

$\boxed{ \pink{ \sf{x \: = \dfrac{ m_1x_2 + m_2x_1}{ m_1 + m_2}}}}$

Apply the formula:

$\implies\sf{x \: = \dfrac{ 3 \times 2 + 4 \times - 2}{ 3 + 4}}$

$\implies\sf{x \: = \dfrac{6-8}{7}}$

$\implies\sf{x \: = \dfrac{-2}{7}}$

Similarly,

$\boxed{ \orange{ \sf{y \: = \dfrac{ m_1y_2 + m_2y_1}{ m_1 + m_2}}}}$

Apply:

$\implies\sf{y \: = \dfrac{ 3 \times -4 + 4 \times - 2}{ 3 + 4}}$

$\implies\sf{y \: = \dfrac{-12-8}{ 7}}$

$\implies\sf{y \: = \dfrac{-20}{ 7}}$

(x,y) = $\sf{(\dfrac{-2}{ 7}), \dfrac{-20}{7}}$