Question

D divides side BC of ∆ABC in the ratio 1:2. Find the length of AD(coordinate geometry)​

Answer 1

Answer:

AD =  $5\sqrt{5} /3$

Step-by-step explanation:

given

b = -1,-2

c= 4,2

ratio given-1;2

m1 = 1

m2=2

now let pt d be (x,y)

x,y = m1x2+m2x1/m1+m2 , m1y2+m2y1/m1+m2

=  1(4) + 2(-1) / 1+2   ,              1 ( 2) + 2(-2) / 1+2

= 4-2 /3                      ,          2-4 /3

2/3 = x                       , y =-2/3

now AD =

using distance formula

AD = $\sqrt{ (2/3)2+ (11/3)2}\\\sqrt{4+121 / 9} \\\\\\\sqrt{125/9} \\5\sqrt{5}/3\\thus , AD= 5\sqrt{5}/3$

Answer 2

Given:

• B = (-1,-2)
• C = (4,2)
• A = (0,3)
• D divides BC of triangle ABC in the ration of 1:2

To Find:

• Length of AD.

Solution:

• Let the ratio 1:2 = m:n
• Let vertices be (x,y)
• ($x_1,y_1$) = (-1,-2)
• $(x_2,y_2)$ = (4,2)
• ($x_3,y_3$) = (0,3)
• (x,y) = ( $\frac{m*x_2+n*x_1}{m+n}$, $\frac{m*y_2+n*y_1}{m+n}$)
• x = $\frac{1*4+2*(-1)}{2+1} = \frac{2}{3}$
• y = $\frac{1*2+2*(-2)}{3} = -\frac{2}{3}$  
• To find the length of AD, we are using the distance formula given by,
• AD = $\sqrt{x^2+y^2}$  
• Substituting the values for "x" and "y"
• AD = $\sqrt{(\frac{2}{3} )^{2}+(\frac{-2}{3})^2 }$  = $\sqrt{\frac{4}{9} +\frac{4}{9} } = \sqrt{\frac{16}{9} } = \frac{4}{3} = 1.3$  
• AD = 1.3 cm.

The length of AD = 1.3 cm