Question

For what value of k are the points (1,1),(3,k) and (-1,4) collinear

Answer 1

Answer.

k = -2

Explanation.

Let the point will be given as

(1,1) , (3,k) , (-1,4)

therefore,

X1 = 1 and Y1 = 1

X2 = 3 and Y2 = k

X3 = -1 and Y3 = 4

Formula if the point are collinear,

X1 ( Y2 - Y3) + X2 ( Y3 - Y1) + X3 ( Y1 - Y2) = 0

1 ( k - 4 ) + 3 ( 4 - 1 ) + (-1) ( 1 - k) = 0

k - 4 + 9 - 1 + k = 0

2k + 4 = 0

k = -2

Therefore,

value of k = -2

Answer 2

$\rm\large\blue{\underline{\underline{ Question : }}}$

For what value of k are the points (1,1),(3,k) and (-1,4) collinear

$\rm\large\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• (1,1) ; (3,k) ; (-1,4)
• These points are Collinear Points.

★ To find,

• The value of k.

★ Let,

• x1 = 1 ; y1 = 1
• x2 = 3 ; y2 = k
• x3 = - 1 ; y3 = 4
• ∆ = 0

By using area of triangle formula

$\tt\red{ 0= \frac {1}{2} | x_{1}(y_{2} - y_{3})+ x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| }$

• Substitute the values.

$\bf\:\implies 0 \times 2 = | 1( k - 4) + 3(4 - 1) +(-1)(1 - k) |$

$\bf\:\implies 0 = | k - 4 + 3(3) - 1 + k |$

$\bf\:\implies0 = | k - 4 + 9 - 1 + k |$

$\bf\:\implies0 = | 2k + 4 |$

$\bf\:\implies 0 - 4 = 2k$

$\bf\:\implies 2k =- 4$

$\bf\:\implies k = \frac{-4}{2}$

$\bf\:\implies k = -2$

$\underline{\boxed{\bf{\purple{ \therefore The\:value\:of\:k = -2}}}}\:\orange{\bigstar}$