Advertisements
Advertisements
Question
For what values of k are the points (8, 1), (3, –2k) and (k, –5) collinear ?
Solution
\[\therefore \frac{1}{2}\left[ 8\left( - 2k + 5 \right) + 3\left( - 5 - 1 \right) + k\left( 1 + 2k \right) \right] = 0\]
\[ \Rightarrow - 16k + 40 - 18 + k + 2 k^2 = 0\]
\[ \Rightarrow 2 k^2 - 15k + 22 = 0\]
\[ \Rightarrow 2 k^2 - 11k - 4k + 22 = 0\]
\[ \Rightarrow k\left( 2k - 11 \right) - 2\left( 2k - 11 \right) = 0\]
\[ \Rightarrow \left( k - 2 \right)\left( 2k - 11 \right) = 0\]
\[ \Rightarrow k - 2 = 0 or 2k - 11 = 0\]
\[ \Rightarrow k = 2 or k = \frac{11}{2}\]
Thus, the values of k are 2 and
\[\frac{11}{2}\]
APPEARS IN
RELATED QUESTIONS
If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)
Find all possible values of x for which the distance between the points
A(x,-1) and B(5,3) is 5 units.
Using the distance formula, show that the given points are collinear:
(6, 9), (0, 1) and (-6, -7)
Find the value of y for which the distance between the points A (3, −1) and B (11, y) is 10 units.
Find the distance between P and Q if P lies on the y - axis and has an ordinate 5 while Q lies on the x - axis and has an abscissa 12 .
P(5 , -8) , Q (2 , -9) and R(2 , 1) are the vertices of a triangle. Find tyhe circumcentre and the circumradius of the triangle.
From the given number line, find d(A, B):
A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.
Find distance CD where C(– 3a, a), D(a, – 2a)