Question

for the value of k are the points a(1, 1) b(3, k) and c(-1,4) collinear​

Answer 1

Correct Question :-

Find the value of k for which the points A(1,1), B(3,k) and C(-1,4) are collinear.

Solution :-

$\bf {\: Area \: of \: triangle \: = \frac{1}{2}( {x}_{1}( {y}_{2} - {y}_{3}) + {x}_{2}( {y}_{3} - {y}_{1}) + {x}_{3}( {y}_{1} - {y}_{2}) }\bf \:$

Here,

• x1 = 1
• x2 = 3
• x3 = -1
• y1 = 1
• y2 = k
• y3 = 4

substituting values in the formula,

= 1/2(1(k-(-1) + 3(4-1) + (-1)(1-k)

since, the points are collinear, the area is equal to 0.

=> 1/2(1(k-(-1) + 3(4-1) + (-1)(1-k) = 0

=> 1/2(1(k+1) + 3(3) -1(1-k) = 0

=> 1/2(k + 1 + 9 - 1 + k) = 0

=> 1/2(2k + 9) = 0

=> 2k + 9 = 0

=> 2k = -9

=> k = -9/2

Answer 2

If the three points are collinear, their area formed will be Zero.

The Entire Solution Is Given In The Above Attachment.

Solving we get K = -3.5