Question

Find the ratio in which the point P(x 2) divides the line segment joining the points A(12 5) and B(4 -3) .Also find the value of x .(2014D)​

Answer 1

Given :-

• Point P(x, 2) divides line segment joining the points A(12, 5) & B(4, -3)

To Find :-

• Ratio in which the point P divides AB.
• Value of x.

Solution :-

Let the ratio in which the point P divides line segment AB be k:1.

Then, According to the section formula, The abscissa of the point P is given by:

⇒ Abscissa of P = (mx₂ + nx₁) / (m + n)

Here, m = k, n = 1; Abscissa of P = x (given)

⇒ x = { k(4) + 1(12) } / (k + 1)

⇒ x = (4k + 12) / (k + 1) ...(i)

Similarly, Using the section formula, Ordinate (y coordinate) of the point P is given by,

⇒ Ordinate of P = (my₂ + ny₁) / (m + n)

Here, m = k, n = 1; Ordinate of P = 1 (given)

⇒ 1 = { k(-3) + 1(5) } / (k + 1)

⇒ k + 1 = -3k + 5

⇒ k + 3k = 5 - 1

⇒ 4k = 4

⇒ k = 1

Substitute k = 1 in eq.(i), we get

⇒ x = { 4(1) + 12 } / (1 + 1)

⇒ x = ( 4 + 12 ) / 2

⇒ x = 16 / 2

⇒ x = 8

We assumed the ratio to be k:1 and found the value of k as 1. Hence, The Point P divides the line segment joining the points A and B in the ratio 1:1 which means P is the midpoint of AB.

Also, The value of x is 8.

Answer 2

Answer:

Given :-

• The point P(x , 2) divides the line segment joining the points A(12 , 5) and B(4 , - 3).

To Find :-

• What is the ratio and the value of x.

Formula Used :-

➦ By using the section formula, the coordinates of the point P are :

$\longmapsto \sf\boxed{\bold{\pink{P =\: \bigg(\dfrac{m{x_2} + n{x_1}}{m + n} , \dfrac{m{y_2} + n{y_1}}{m + n}\bigg)}}}$

Solution :-

$\mapsto$ Given points are :

• A(12 , 5)
• B(4 , - 3)

Let, P(x , 2) divides the line segment AB in the ratio of k : 1

Given :

• m = k
• n = 1
• P = x
• x₁ = 12
• x₂ = 4

Now, by comparing coordinates of the point P, we have :

$\implies \sf P =\: \bigg(\dfrac{m{x_2} + n{x_1}}{m + n}\bigg)$

$\implies \sf x =\: \bigg(\dfrac{k \times 4 + 1 \times 12}{k + 1}\bigg)$

$\implies \sf x =\: \bigg(\dfrac{4k + 12}{k + 1}\bigg)$

$\implies \sf\bold{\purple{x =\: \bigg(\dfrac{4k + 12}{k + 1}\bigg)}}$

Now, again we have :

Given :

• m = k
• n = 1
• P = 1
• y₁ = 5
• y₂ = - 3

Then, similarly we get,

$\implies \sf P =\: \bigg(\dfrac{m{y_2} + n{y_1}}{m + n}\bigg)$

$\implies \sf 1 =\: \bigg(\dfrac{k \times (- 3) + 1 \times 5}{k + 1}\bigg)$

$\implies \sf 1 =\: \bigg(\dfrac{- 3k + 5}{k + 1}\bigg)$

By doing cross multiplication we get ;

$\implies \sf k + 1 =\: - 3k + 5$

$\implies \sf k + 3k =\: 5 - 1$

$\implies \sf 4k =\: 4$

$\implies \sf k =\: \dfrac{\cancel{4}}{\cancel{4}}$

$\implies \sf k =\: \dfrac{1}{1}$

$\implies \sf\bold{\green{k =\: 1}}$

Now, we have to find the value of x :

So, we have the value of k is 1,

Now, by putting the value of k in this equation we get,

$\implies \sf x =\: \bigg(\dfrac{4k + 12}{k + 1}\bigg)$

$\implies \sf x =\: \bigg(\dfrac{4(1) + 12}{1 + 1}\bigg)$

$\implies \sf x =\: \bigg(\dfrac{4 \times 1 + 12}{2}\bigg)$

$\implies \sf x =\: \bigg(\dfrac{4 + 12}{2}\bigg)$

$\implies \sf x =\: \dfrac{\cancel{16}}{\cancel{2}}$

$\implies \sf\bold{\red{x =\: 8}}$

$\therefore$ The ratio in which the point P(x , 2) divides the line segment joining the points A (12 , 5) and B(4 , - 3) is 1 : 1

Because we let the ratio is k : 1 and then we get the value of k is 1, so 1 : 1 is the ratio.

$\therefore$ The value of x is 8.

$\rule{300}{2}$

IMPORTANT FORMULA RELATED TO CO-ORDINATOR GEOMETRY :-

$\clubsuit$ Distance Formula :

$\leadsto \sf\boxed{\bold{\pink{d =\: \sqrt{{(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2}}}}}$

where,

• d = Distance
• (x₁ + x₁) = Coordinates of the first point
• (x₂ + y₂) = Coordinates of the second point

$\clubsuit$ Mid-point Formula :

$\leadsto \sf\boxed{\bold{\pink{Mid-point =\: \bigg \lgroup \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg\rgroup}}}$

where,

• (x₁ + x₂) = Coordinates of the x-axis
• (y₁ + y₂) = Coordinates of the y-axis

$\clubsuit$ Centroid Formula :

$\leadsto \sf\boxed{\bold{\pink{C =\: \bigg(\dfrac{x_1 + x_2 + x_3}{3} , \dfrac{y_1 + y_2 + y_3}{3}\bigg)}}}$

where,

• C = Centroid of the triangle
• x₁ , x₂ , x₃ = x-coordinates of the vertices of the triangle
• y₁ , y₂ , y₃ = y-coordinates of the vertices of the triangle