Question

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Answer 1

Question: Find the ratio in which the linesegment joining the points (-3,10) and (6,-8) is divided by (-1,6).

Solution:

Let us name the linesegment to be AB.

⇒ A(-3, 10)

⇒ B(6, -8)

The point (-1,6) divides the linesegment, let's name it D.

We know that the ratio in which a linesegment is divided can be found out by using the Section Formula, assuming that the line is divided in the ratio k:1.

When the line is divided in the ratio k:1, we'll use the formula;

$\sf \ D(-1,6) = \left(\dfrac{kx_2 + x_1}{k + 1}, \dfrac{ky_2 + y_1}{k + 1}\right)$

Substituting the coordinates we get,

$\sf \ D(-1,6) = \left(\dfrac{kx_2 + x_1}{k + 1}, \dfrac{ky_2 + y_1}{k + 1}\right)$

$\sf \ D(-1,6) = \left(\dfrac{k(6) + (-3)}{k + 1}, \dfrac{k(-8) + 10}{k + 1}\right)$

$\sf \ D(-1,6) = \left(\dfrac{6k - 3}{k + 1}, \dfrac{-8k + 10}{k + 1}\right)$

We know that the 'x' coordinate and the 'y' coordinate are equal to (kx₂ + x₁/k + 1) and (ky₂ + y₁/k + 1) respectively. So, we equate the 'x' coordinate with {6k - 3}/{k + 1}

$\Longrightarrow \sf -1 = \left(\dfrac{6k - 3}{k + 1}\right)$

$\Longrightarrow \sf -k + -1 = 6k -3$

$\Longrightarrow \sf -k - 6k = -3 + 1$

$\Longrightarrow \sf -7k = -2$

$\Longrightarrow \sf 7k = 2$

$\Longrightarrow \sf k = \dfrac{2}{7}$

$\Longrightarrow \sf k:1 = 2:7$

∴ The line is divided in the ratio 2:7

Answer 2

"ɪ ᴅᴏɴ'ᴛ ᴄᴀʀᴇ ᴡʜᴏ ɪ ʜᴀᴠᴇ ᴛᴏ ғɪɢʜᴛ!

ɪғ ʜᴇ ʀɪᴘs ᴍʏ ᴀʀᴍs ᴏᴜᴛ, ɪ'ʟʟ ᴋɪᴄᴋ ʜɪᴍ ᴛᴏ ᴅᴇᴀᴛʜ!

ɪғ ʜᴇ ʀɪᴘs ᴍʏ ʟᴇɢs ᴏғғ, ɪ'ʟʟ ʙɪᴛᴇ ʜɪᴍ ᴛᴏ ᴅᴇᴀᴛʜ!

ɪғ ʜᴇ ʀɪᴘs ᴍʏ ʜᴇᴀᴅ ᴏғғ, ɪ'ʟʟ sᴛᴀʀᴇ ʜɪᴍ ᴛᴏ ᴅᴇᴀᴛʜ!

ᴀɴᴅ ɪғ ʜᴇ ɢᴏᴜɢᴇs ᴏᴜᴛ ᴍʏ ᴇʏᴇs, ɪ'ʟʟ ᴄᴜʀsᴇ ʜɪᴍ ғʀᴏᴍ ᴛʜᴇ ɢʀᴀᴠᴇ!

ᴇᴠᴇɴ ɪғ ɪ'ᴍ ᴛᴏʀɴ ᴛᴏ sʜʀᴇᴅs, ɪ'ᴍ ᴛᴀᴋɪɴɢ sᴀsᴜᴋᴇ ʙᴀᴄᴋ ғʀᴏᴍ ᴏʀᴏᴄʜɪᴍᴀʀᴜ!"

—ᴜᴢᴜᴍᴀᴋɪ ɴᴀʀᴜᴛᴏ