Question

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Answer 1

Given Question :-

Point P divides the line segment joining the points A and B in the ratio k : 1. If the coordinates of point A (5, 6) and B (12, - 8), and P lies on x - axis. Find the value of k.

Solution :-

Given that,

• The coordinates of point A (5, 6) and B (12, - 8), and P lies on x - axis.

So,

• Let assume that coordinates of P be (x, 0).

Further,

• P divides AB in the ratio k : 1.

We know,

Section Formula,

Let us consider a line segment joining the points A and B and let C (x, y) be any point which divides the line segment AB in the ratio m : n internally then coordinates of C is

$\boxed{\bf \:( x, y) = \bigg (\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} \bigg)}$

Here,

$\rm :\longmapsto\:x_1 = 5,y_1 = 6,x_2= 12,y_2= - 8 ,m =k ,n =1$

So, on substituting the values, we get

$\rm :\longmapsto\:( x, 0) = \bigg (\dfrac{12k + 5}{k + 1} , \dfrac{ - 8k + 6}{k + 1} \bigg)$

On comparing, we get

$\rm :\longmapsto\:\dfrac{ - 8k + 6}{k + 1} = 0$

$\rm :\longmapsto\: - 8k + 6 = 0$

$\rm :\longmapsto\: - 8k = - 6$

$\rm :\implies\:k = \dfrac{3}{4}$

Additional Information :-

Distance Formula :-

Let us consider a line segment joining the points A and B then distance between A and B is

$\rm :\longmapsto\: \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

Midpoint Formula :-

Let us consider a line segment joining the points A and B and let C be the midpoint of AB then coordinates of C is

$\rm :\longmapsto\: \:( x, y) = (\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} )$

Area of triangle :-

$\rm \ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$