Question

if a (4,1),b(-2,3)and c(0,5)and vertices of triangle abc and ad is its median then if P(2/3 ,3) is a point on ad ,then the ratio ap:dp is​

Answer 1

Given info : if a (4,1) , b(-2,3) and c(0,5)and vertices of triangle abc and ad is its median.

To find : if P(2/3 ,3) is a point on ad ,then the ratio ap : dp is​

solution : ad is median on bc. so point d is the midpoint of line bc.

using midpoint section formula, $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

so, point d = [(-2 + 0)/2 , (3 + 5)/2] = (-1 , 4)

now a(4,1) ------------P(2/3 , 3)-------------d(-1,4)

let ratio of ap : dp is m : n

using section formula, $\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{2}\right)$

so, 2/3 = (m × -1 + n × 4)/(m + n)

⇒ 2(m + n) = 3(-m + 4n)

⇒ 2m + 2n = -3m + 12n

⇒ 5m = 10n

⇒ m/n = 2/1

Hence, the ratio of ap : dp is 2 : 1 (internally)