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

Step-by-step explanation:

Given: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.

To find:Ratio AP:DP is ?

Solution:

Tip:Section formula

if $A(x_1,y_1)$ and $B(x_2,y_2)$ are the end points of a line segment and point P divides AB in m:n, then coordinates of P $\left(\bold{\red{\frac{mx_1+nx_2}{m+n},\frac{my_1+ny_2}{m+n}}}\right)$.

Step 1: Find mid-point of BC

B(-2,3) and C(0,5)

$D\left(\frac{-2+0}{2},\frac{3+5}{2}\right)$

$D\left(\frac{-2}{2},\frac{8}{2}\right)$

D(-1,4)

Step 2: Apply section formula,let say P divides AD in m:n

Applying for x coordinate of P

$\frac{2}{3}=\frac{4m-n}{m+n}$

Simplify

$2m+2n=12m-3n\\\\-10m=-5n\\\\\frac{m}{n}=\frac{-5}{-10}\\\\\frac{m}{n}=\frac{1}{2}\\\\\frac{n}{m}=\frac{2}{1}\\\\\bold{\red{\frac{AP}{DP}=\frac{2}{1}}}$

Final answer:

AP:DP=2:1

Hope it helps you.

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