Question

For what value of K do the points (-1,4),(-3,8) and (-K+1,3K) lie on a straight line .

Answer 1

you can check this by determinant method.. which I have solved.. or u may consider the three points to b the coordinate of the triangle and find the area.. there after equate it with 0 as 3 collonear points will have area of 0 units

Answer 2

Concept

When three points are collinear then the condition to fulfill collinearity is given by the determinant form which is as follows,

$\left[\begin{array}{ccc}x&y&1\\p&q&1\\s&t&1\end{array}\right] = 0$

where (x, y) is the first point, (p, q) is the second point and (s, t) is the third point on the line. Which means that the determinant should be zero. The formula is similar to the area of the triangle which is obvious that when the points are on the same straight line then the area enclosed by them will be zero.

Given

The following three points on the lines is given as,

(x, y) = (-1, 4)

(p, q) = (-3, 8)

(s, t) = (-k+1, 3k)

Find

We have to calculate the value of k using the above condition.

Solution

$\left[\begin{array}{ccc}-1&4&1\\-3&8&1\\-k+1&3k&1\end{array}\right] = 0\\\\-1(8 - 3k ) -4(-3 + k -1) +1(-9k + 8k -8) =0\\3k - 8 - 4k +16 - k -8 =0\\2k = 0\\k = 0$

Hence the value of k is 0.

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