Question 30 Section Formula Exercise 11
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Answer:
Solution:
In coordinate geometry, the Section formula is used to determine the internal or external ratio at which a line segment is divided by a point.
Let the ratio that the point (5,4) divide the line segment joining the points (2,1) and (7,6) be m:n,
Here x1 = 2 , y1 = 1 , x2 = 7, y2 = 6, x = 5, y = 4
By section formula,
x=\frac{\left(mx_2+nx_1\right)}{(m+n)}\\5=\frac{\left(m\times7+n\times2\right)}{(m+n)}\\5=\frac{\left(7m+2n\right)}{(m+n)}\\5(m+n)=7m+2n\\ 5m+5n=7m+2n\\ 5m-7m=2n-5n\\-2m=-3n\\ \frac{m}{n}=\frac{-3}{-2}=\frac{3}{2}
Hence the ratio m:n is 3:2.
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Text Solution
Solution : Let the required ratio be k : 1. <br> Then, by the section formula, the coordinates of P are <br> `P((4k - 3)/(k+1), (-9k+5)/(k+1))` <br> `therefore (4k-3)/(k+1) = 2 "and" (-9k+5)/(k+1) =-5 " "[because P(2, -5) "is given"]` <br> `rArr 4k-3 = 2k+2"and" -9k+5 =-5k-5` <br> `rArr 2k =5 "and" 4k = 10` <br> `rArr k = (5)/(2)` in each case. <br> So, the required ratio is `(5)/(2):1,"which is" 5:2.` <br> Hence, P divides AB in the ratio 5:2.