Question

find the ratio in which the line segment joining the points A(3,8) and B(-9,3) is divided by the Y-axis.​

Answer 1

$\huge \pink {answeR}$

here coordinates of dividing point is

(0,y)

p(x,y) = { mx2 + nx1/m+n , my2 + ny1/m+n}

p(0,y) = { m(-9)+n(3)/m+n , m(3)+n(8)/m+n}

p(0,y) = { -9m+3n/m+n , 3m+8n/m+n }

compare both side

0 = -9m+3n/m+n

0(m+n) = -9m+3n

0 = -9m+3n

9m = 3n

m/n = 3/9

m/n = 1/3

m:n = 1:3

so the ratio is 1:3

Answer 2

$\sf{Answer}$

Step by step explanation:-

Given :-

The line segment joining points A=(3,8) B=(-9 ,3) Is divided by Y-axis

To find :-

The ratio of line segment ?

Concept to know :-

The line joining the points (x1 ,y1) and (x2,y2) is divided by Y-axis in ratio

$\sf\dfrac{-x_1}{x_2}$

Solution:-

A(3,8) and B(-9,3)

So,

x1 = 3

x2 = -9

y1 = 8

y2 = 3

So, the ratio of y-axis $\sf\dfrac{-x_1}{x_2}$

= $\sf\dfrac{-3}{-9}$

= $\sf\dfrac{3}{9}$

= $\sf\dfrac{1}{3}$

So, the ratio of line segment joining points A(3,8) and B(-9,3) by Y-axis is $\sf\dfrac{1}{3}$

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Know more :-

The distance between points P = (x1,y1) and (x2,y2) is

$\sf\sqrt{(x_1 - x_2)^2 +(y_1 - y_2)^2}$

If P is the midpoint of AP then m=n the cordinates of middle points of AB are

$\sf\dfrac{x_1 + x_2}{2}$ , $\sf\dfrac{y_1+y_2}{2}$

Centroid formula:-

$\sf\dfrac{x_1+x_2+x_3}{3}$,$\sf\dfrac{y_1+y_2+y_3}{3}$

Section formula for Internal division :-

$\sf\dfrac{mx_2+nx_1}{m+n}$, $\sf\dfrac{my_2+ny_1}{m+n}$

Where m, n is the ratio which divides internally

Section formula for External division

$\sf\dfrac{mx_2-nx_1}{m-n}$, $\sf\dfrac{my_2-ny_1}{m-n}$

Where m, n is the ratio which divides externally