Question

Calculate the ratio in which the line jong
A(-4, 2) and B3, 6) is divided by som
P(x, 3). Also, find. length of AP​

Answer 1

Answer:

let the ratio k:1 divides the line a,b

(x,3)=)3k-4/k+4,6k+2/k+1)

6k+2=3k+3

3k=1

k=1/3

so the ratio is 1:3

Answer 2

✪ᴀɴsᴡᴇʀ✪

• P divides AB in the ratio 1:3.
• $AP = \frac{\sqrt{65}}{4}$

★ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ★

Let us assume that P(x, 3) divides line joining A(-4, 2) and B(3,6) in the ratio k:1, then by section formula

$(x, 3)=\left(\frac{3k-4}{k+1},\frac{6k+2}{k+1}\right)$

$x =\frac{3k-4}{k+1},3=\frac{6k+2}{k+1}$

$x =\frac{3k-4}{k+1},3k+3=6k+2$

$x =\frac{3k-4}{k+1},3-2=3k$

$x =\frac{3k-4}{k+1},k=\frac{1}{3}$

$x =\frac{3\frac{1}{3}-4}{\frac{1}{3}+1},k=\frac{1}{3}$

$x =\frac{3-12}{1+3},k=\frac{1}{3}$

$x =-\frac{9}{4},k=\frac{1}{3}$

Thus P divides AB in the ratio 1:3.

Also

$AP = \sqrt{(x+4)^2 + (3-2)^2 }$

$AP = \sqrt{(-\frac{9}{4}+4)^2 + 1 }$

$AP = \sqrt{(\frac{-9+16}{4})^2 + 1 }$

$AP = \sqrt{(\frac{7}{4}^2 + 1 }$

$AP = \sqrt{(\frac{49+16}{4^2} }$

$AP = \sqrt{\frac{65}{4^2} }$

$AP = \frac{\sqrt{65}}{4}$