What ratio is the segment joining the point 4 3 and 5 4 divided by x-axis?

In what ratio does the x axis divide the join of A2, 3 and B 5, 6? a 2 : 3 b 3 : 5 c 1 : 2 d 2 : 1

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$\frac{6k-3}{k+1}=0$

$6k-3=0$

$k=\frac{3}{6}=\frac{1}{2}$

Thus, x-axis divides the line segment joining the points (2, –3) and (5,6) in the ratio 1:2.

Question 5 Coordinated Geometry - Exercise 7.3

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Answer:

Let the ratio in which x-axis divides the line segment joining (–4, –6) and (–1, 7) = 1: k.

Then,

x-coordinate becomes, \frac{\left(-1-4k\right)}{(k+1)}

y-coordinate becomes, \frac{\left(7-6k\right)}{(k+1)}

Since P lies on x-axis, y coordinate = 0

\frac{\left(7-6k\right)}{(k+1)}=0\\ 7-6k=0\\ k=\frac{7}{6}

Therefore, the point of division divides the line segment in the ratio 6 : 7.

Now, m1 = 6 and m2 = 7

By using the section formula,

x=\frac{\left(m_1x_2+m_2x_2\right)}{(m_1+m_2)}=\frac{\left[6(-1)+7(-4)\right]}{(6+7)}=\frac{\left(-6-28\right)}{13}=-\frac{34}{13}\\ So,\ now\\ y=\frac{\left[6(7)+7(-6)\right]}{(6+7)}=\frac{\left(42-42\right)}{13}=0

Hence, the coordinates of P are (-34/13, 0)

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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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Page 2

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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