# find in what ratio is the line joining (3,4) and (5,-7) cut by the co-ordinates axis

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The general equation for a line is y = mx+c

Our line goes through (3,4) and (5,-7)

So we can write;

4=m*3+c------------(1)

-7=m*5+c------------(2)

(1)-(2)

11 = -2m

m= -11/2

From (1) we can get 4=(-11/2)*3+c

c =41/2

Equation of the line joining (3,4) and (5,-7) is y=(-11/2)x+41/2

Since y values of two points changes 4 to -7 the line will cut x axis.

When y=0 then x=(41/2)(2/11) = 41/11

So the cutting point is (41/11,0)

Let legth AB=length (3,4) to ((41/11,0)

BC= length (5,-7) to ((41/11,0)

AB=sqrt[(3-41/11)^2+(4-0)^2] = sqrt[2000/121]

BC=sqrt[(5-41/11)^2+(-7-0)^2]= sqrt[6125/121]

So the ratio is AB/BC =sqrt[2000/121]/sqrt[6125/121]

=sqrt[2000/6125]

= 4/7

**The the cordinate axis cut the line to the ratio of 4:7**

To find the ratio of the line cut by the x-axis, we need to use the distance formula `d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}` between the two points `(x_1,y_1)` and `(x_2,y_2)`.

The equation of the line between the end points is found by getting the slope of the line.

`m={-7-4}/{5-3}`

`=-11/2`

The equation of the line is then found using `y=mx+b` and solving for b.

`4=-11/2\cdot 3+b`

`b={8+33}/2`

`b=41/2`

so the equation of the line is `y=-11/2 x +41/2` .

The x-intercept of this line is at `y=0`. That is, when `0=-11/2 x+41/2`

which is at `x=41/11`.

Consider the three points `A=(3,4)`, `B=(41/11,0)` and `C=(5,-7)`. The point B is the x-intercept of the line.

The distance from A to B is

`d_{AB}=\sqrt{(3-41/11)^2+(0-4)^2}`

`=\sqrt{64/121+16}`

`=\sqrt{2000}/11`

and the distance from B to C is

`d_{BC}=\sqrt{(5-41/11)^2+(-7-0)^2}`

`=\sqrt{196/121+49}`

`=\sqrt{6125}/11`

**This means that the ratio from AB to BC after cancelling out the common factor of 11 is**

**`\sqrt{2000/6125}` **

**`=\sqrt{16/49}` **

**`=4/7` **