Question

in what ratio is the line segment joining A(2,-3)&B(5,6) divided by x-axis also find the coordinate of the point of division.

Answer 1

Answer:

The point of division is (3,0)

Step-by-step explanation:

Concept:

The co ordinates of the point which divides the line segment joining $(x_1,y_1)$ and $(x_2,y_2)$ internally in the ratio m:n is

$\displaystyle(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

Let P be the point on x axis which divides line joining A(2,-3) and B(5,6) internally in the ratio m:n

Then, the coordinates P is

$\displaystyle(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

$\displaystyle(\frac{m(5)+n(2)}{m+n},\frac{m(6)+n(-3)}{m+n})$

$\displaystyle(\frac{5m+2n}{m+n},\frac{6m-3n}{m+n})$

since P lies on x axis, y coordinate of P is 0

$\displaystyle\frac{6m-3n}{m+n}=0$

$6m-3n=0$

$6m=3n$

$\frac{m}{n}=\frac{3}{6}$

$\frac{m}{n}=\frac{1}{2}$

Hence,the x axis divides line joining A and B internally in the ratio 1:2

The coordinates of P is

$\displaystyle(\frac{5(1)+2(2)}{1+2},\frac{6(1)-3(2)}{1+2})$

$(\frac{9}{3},\frac{0}{3})$

$(3,0)$

Find more:

In what ratio does the x- axis divide the line segment joining A(5,6) and B (2,-8)?​

https://brainly.in/question/13579888

Answer 2

We know that

x = M1x2 +M2x1/M1+ M2

And

y = M1y2 + M2y1/M1 +M2

But we know that y=0

So

0 =6M1-3M2/M1 + M2

0 = 6M1 -3M2

-6M1 = -3M2

M1/M2 = 1/2

The ratio is 1 : 2

x = 1*5 +2*2/1+3 = 5+4 /3 =9/3 =3 The coordinate is (3,0)