Question

The area of the Triangle formed by the Points (1, 2), (k, 5), (7, 11) is 0, then the values of k is​

Answer 1

Step-by-step explanation:

Given area of triangle whose vertices are (1,−1),(−4,2k) and (−k,−5) is 24 sq units

We know that the area of triangle =

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

∴24=

2

1

[1(2k−(−5))−4(−5−(−1))+(−k)(−1−2k)]

⇒48=[2k+5+(−4)(−4)+k(1+2k)]

⇒48=[2k+5+16+k+2k

2

]

⇒2k

2

+3k+5+16−48=0

⇒2k

2

+3k−27=0

⇒2k

2

+9k−6k−27=0

⇒k(2k+9)−3(2k+9)=0

⇒(k−3)(2k+9)=0

⇒k−3=0or2k+9=0

⇒k=3ork=

2

−9

Answer 2

Given: The area of the Triangle formed by the Points (1, 2), (k, 5), (7, 11) is 0

We know that the area of triangle = 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

∴0=1/2[1(5-11)+k(11-2)+7(2-5)]

0×2=1(-6)+k(9)+7(-3)

-6+9k-21=0

9k-27=0

9k=27

k=27/9

k=9/3

k=3