Question

for what value of k the point p(-5,7) q (k,5)and r (0,2) r collinear​

Answer 1

Answer:

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Step-by-step explanation:

so, x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)=0

Take (7,−2) as (x

1

,y

1

), (5,1) as (x

2

,y

2

) and (3,k) as (x

3

,y

3

), we have

7(1−k)+5(k+2)+3(−2−1)=0

⇒ 7−7k+5k+10−9=0

⇒ 2k=8

⇒ k=4