Question

Find the value of k, if (7. k) lies on the line passing through the points
(-5,2), (3,6).​

Answer 1

Answer:

the answer is 11.5

Explanation:

let A(-5,2),B(3,6),C(7,k)

eqn : (x1(y2-y3)+x2(y3-y1)+x3(y1-y2))=0

then-5 is x1,2 is y1,3is x2,

6 is y2,7is x3 ,k is y3

so,

(-5(6-k)+3(k+5)+7(-5-6))=0

-30+5k +3k+15-77=0

-30-77+15+8k=0

-92+8k=0

8k=92

k=92/8=11.5

Answer 2

Concept:

Area of triangle if the points are collinear is equal to 0.

Also, the area of triangle given its vertices is:

Area = A = 1 / 2  | x₁ ( y₂ - y₃ ) + x₂ ( y₃ - y₁ ) + x₃ ( y₁ - y₂ ) |

Given:

We are given three points:

(7, k), (-5, 2) and (3, 6).

Find:

We need to find the value of k.

Solution:

Area = A = 1 / 2  | x₁ ( y₂ - y₃ ) + x₂ ( y₃ - y₁ ) + x₃ ( y₁ - y₂ ) | = 0

Substituting the values of the points:

1 / 2  | 7 ( 2 - 6 ) + (-5) ( 6 - k ) + 3 ( k - 2 ) | = 0

| 7(-4) + (-5)(6 - k) + 3(k - 2) |= 0

| -28 -30 +5k + 3k - 6 | = 0

| 8 k - 64 | = 0

8k - 64 = 0    or    64 - 8k = 0

8k = 64

k = 8.

Therefore, the value of k is 8.

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