In what ratio is the line joining the points (4,2) and (3,5) divided by the x axis

In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.

Let the point P (x, 0) on x-axis divides the line segment joining A (4, 3) and B (2, -6) in the ratio k: 1.
Using section formula, we have:

`0=(-6k+3)/(k+1)`

`0=-6k+3`

`k=1/2`

Thus, the required ratio is 1: 2.
Also, we have:

`x=(2k+4)/(k+1)`

`x=(2xx1/2+4)/(1/2+1)`

`x=10/3`

Thus, the required co-ordinates of the point of intersection are `(10/3,0)`

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Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the coordinates of the point of intersection.

`0=(3k-4)/(k+1)`

`3k=4`

`k=4/3` ..............(1)

`y=(0+7)/(k+1)`

`y=7/(4/3+1)` (from eq. 1)

`y=3`

Hence, the required is 4:3 and the required point is S(0, 3)

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Let the point P which is on the x-axis, divide the line segment joining the points A(4, 2) and B(3, -5) in the ratio of m : n. Let the coordinates of P be (x, 0).

By section formula,

y-coordinate = $my2+ny1m+n\dfrac{my2 + ny1}{m + n}$

$⇒0=m×(−5)+n×2m+n⇒0=−5m+2n⇒5m=2n⇒mn=25⇒m:n=2:5.\Rightarrow 0 = \dfrac{m \times (-5) + n \times 2}{m + n} \\[1em] \Rightarrow 0 = -5m + 2n \\[1em] \Rightarrow 5m = 2n \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{5} \\[1em] \Rightarrow m : n = 2 : 5.$

Putting value of m : n in section-formula for x-coordinate,

x-coordinate = $mx2+nx1m+n\dfrac{mx2 + nx1}{m+ n}$

$\dfrac{2 \times 3 + 5 \times 4}{2 + 5} \\[1em] = \dfrac{6 + 20}{7} \\[1em] = \dfrac{26}{7}.$

Hence, coordinates of P are $(267,0)(\dfrac{26}{7}, 0)$ and 2 : 5 is the ratio in which the line joining the points (4, 2) and (3, -5) is divided by the x-axis.