 Text Solution Answer : `2:3` Solution : Let the required ratio be k: 1. <br> Then, the point `P((8k +3)/(k+1), (9k-1)/(k+1))` lies on the line x-y-2=0. <br> `therefore (8k+3)/(k +1) - (9k-1)/(k +1)-2 = 0 rArr (8k+3) - (9k-1) -2 (k+1) = 0` <br> `rArr (8k-9k -2k) + (3 +1 -2) = 0` <br> `rArr 3k = 2 rArr k = (2)/(3).` <br> Hence, the required ratio is `((2)/(3) : 1), i.e., 2:3.` Open in App Suggest Corrections
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 line x-y-2 = 0 divide the line **segment joining the points (3,-1) and (8,9) in the ratio m:n at the point P(x,y)** Here x1 = 3, y1 = -1 x2 = 8 y2 = 9 By section formula, x=\frac{\left(mx_2+nx_1\right)}{(m+n)}\\x=\frac{\left(m\times8+n\times3\right)}{(m+n)}\\x=\frac{\left(8m+3n\right)}{(m+n)}\dots.(i) By section formula, y=\frac{\left(my_2+ny_1\right)}{(m+n)}\\y=\frac{\left(m\times9+n\times-1\right)}{(m+n)}\\y=\frac{\left(9m-n\right)}{(m+n)}\dots.(ii) Since the point P(x,y) lies on the line x-y-2 = 0, eqn (i) and (ii) will satify the equation x-y-2 = 0 …(iii) Substitute (i) and (ii) in (iii), \frac{\left(8m+3n\right)}{(m+n)}-\frac{\left(9m-n\right)}{(m+n)}-2=0\\\frac{\left(8m+3n\right)}{(m+n)}-\frac{\left(9m-n\right)}{(m+n)}-2\frac{(m+n)}{(m+n)}=0\\ 8m+3n-(9m-n)-2(m+n)=0\\8m+3n-9m+n-2m-2n=0\\-3m+2n=0\\-3m=-2n\\ \frac{m}{n}=\frac{-2}{-3}=\frac{2}{3} Hence the ratio m:n is 2:3. Substitute m and n in (i), x=\frac{\left(8m+3n\right)}{(m+n)}\\x=\frac{\left(8\times2+3\times3\right)}{(2+3)}\\x=\frac{\left(16+9\right)}{5}\\x=\frac{25}{5}=5 Substitute m and n in (ii), y=\frac{\left(9m-n\right)}{(m+n)}\\y=\frac{\left(9\times2-3\right)}{(2+3)}\\y=\frac{\left(18-3\right)}{(2+3)}\\y=\frac{15}{5}=3 Hence the co-ordinates of P are (5,3). |
What ratio does line XY 2 0 divide the line segment joining the points a 3 1 and B 8 9 CBSE 2007?
Related Posts
Copyright © 2024 SignalDuo Inc.
