In what ratio is the line segment joining the points (5:8 and 7 3 divided by the x axis and y-axis)

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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Let the line segment A(5, -6) and B(-1, -4) is divided at point P(0, y) by y-axis in ratio m:n

Here, (x, y) = (0, y); (x1, y1) = (5, -6) and (x2, y2) = (-1, -4)

So,

⇒ 0 = -m + 5n

⇒ m = 5n

Hence, the ratio is 5:1 and the division is internal.

Now,

Hence, the coordinates of the point of division is

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Text Solution

Solution : Let X-axis divides the join of points (8,5) and (-3, -7) in the ratio `k:1`. <br> We know that at X-axis <br> ` " "y =0` <br> ` rArr" "(k*y_(2)+1*y_(1))/(k+1)=0` <br> `rArr" "(k(-7)+ 1(5))/(k+1)=0` <br> ` rArr" "-7k+5=0` <br> ` rArr" "k=(5)/(7)` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NTN_MATH_X_C07_S01_032_S01.png" width="80%"> <br> `therefore` Required ratio= 5: 7