Question

Find the value of k for which the points (7,-2), (5,1)
and (3,k)

Answer 1

$\huge \mathfrak \fcolorbox{purple}{lavender} {★Answer★}$

ll4 is your correct answerll

sɪɴᴄᴇ ᴘᴏɪɴᴛs ᴀʀᴇ ᴄᴏʟʟɪɴᴇᴀʀ,

↝sᴏ, x1 (ʏ2−ʏ3)+x2(ʏ3−ʏ1)+x3(ʏ1−ʏ2)=0

↝ᴛᴀᴋᴇ (7,−2) ᴀꜱ (x1,ʏ1), (5,1) ᴀꜱ (x2,ʏ2) ᴀɴᴅ (3,ᴋ) ᴀꜱ (x3,ʏ3), ᴡᴇ ʜᴀᴠᴇ

⇒7(1−к)+5(к+2)+3(−2−1)=0

⇒ 7−7k+5k+10−9=0

⇒ 2k=8

⇒ k=4

Answer 2

Answer:

4

Step-by-step explanation:

Correct Question :-

Find the value of k for which the points (7,-2), (5,1)  and (3,k) are collinear.

Given ,

A = (7 , - 2)

B = (5 , 1)

C = (3 , k)

A , B , C are collinear

To Find :-

Value of 'k'

How To Do :-

As they said that A , B , C  are collinear , so we need to equate that area of the triangle to '0' and we need to substitute the value of the co-ordinates of A , B , C in that formula and we need to kind the value of 'k'

Formula Required :-

Area of the triangle :-

$=\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Solution :-

A = (7 , - 2)

Let ,

x_1 = 7 , y_1 = - 2

B = (5 , 1)

Let ,

x_2 = 5 , y_2 = 1

C = (3 , k)

Let,

x_3 = 3 , y_3 = k

Area of the triangle be '0'

Substituting in the formula :-

$0=\dfrac{1}{2}|7(1-k)+5(k-(-2))+3(-2-1)|$

$0= \dfrac{1}{2}|7(1)+7(-k)+5(k+2)+3(-3)|$

$0=\dfrac{1}{2}|7-7k+5(k)+5(2)-9|$

$0=\dfrac{1}{2}|7-7k+5k+10-9|$

$0=\dfrac{1}{2}|7-2k+1|$

$0=\dfrac{1}{2}|8-2k|$

$0\times 2=|8-2k|$

0 = |8 - 2k|

0 = 8 - 2k

-8 = -2k

-8/-2 = k

4 = k

∴ Value of 'k' = 4