मराठी

Find the Value of K If Points A(K, 3), B(6, −2) and C(−3, 4) Are Collinear. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the value of k if points A(k, 3), B(6, −2) and C(−3, 4) are collinear.

 
थोडक्यात उत्तर

उत्तर

The formula for the area ‘A’ encompassed by three points(x1 , y1 ) ,  (x2 , y2 )   and (x3 , y3 )  is given by the formula,

\[∆ = \frac{1}{2}\left| \left( x_1 y_2 + x_2 y_3 + x_3 y_1 \right) - \left( x_2 y_1 + x_3 y_2 + x_1 y_3 \right) \right|\]

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(k, 3), B(6, −2) and C(3, 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

\[∆ = \frac{1}{2}\left| \left( k\left( - 2 \right) + 6 \times 4 + \left( - 3 \right) \times 3 \right) - \left( 6 \times 3 + \left( - 3 \right)\left( - 2 \right) + k \times 4 \right) \right|\]

\[ 0 = \frac{1}{2}\left| \left( - 2k + 24 - 9 \right) - \left( 18 + 6 + 4k \right) \right|\]

\[ 0 = \frac{1}{2}\left| - 2k + 15 - 24 - 4k \right|\]

\[ 0 = \frac{1}{2}\left| - 6k - 9 \right|\]

\[ 0 = - 6k - 9\]

\[ k = - \frac{9}{6} = - \frac{3}{2}\]

Hence the value of ‘k’ for which the given points are collinear is `(k = - 3 /2)`.

 

 
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Co-Ordinate Geometry - Exercise 6.5 [पृष्ठ ५४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 10
पाठ 6 Co-Ordinate Geometry
Exercise 6.5 | Q 15 | पृष्ठ ५४

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Find the distance between the following pair of points:

(a, 0) and (0, b)


Find the third vertex of a triangle, if two of its vertices are at (−3, 1) and (0, −2) and the centroid is at the origin.

 

 

Find the points of trisection of the line segment joining the points:

5, −6 and (−7, 5),


Find the points of trisection of the line segment joining the points:

(2, -2) and (-7, 4).


The line segment joining the points P(3, 3) and Q(6, -6) is trisected at the points A and B such that Ais nearer to P. If A also lies on the line given by 2x + y + k = 0, find the value of k.


Show that the points A (1, 0), B (5, 3), C (2, 7) and D (−2, 4) are the vertices of a parallelogram.


The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x - y + k = 0. Find the value of k.


Find the points on the y-axis which is equidistant form the points A(6,5)  and B(- 4,3) 


Find the ratio which the line segment joining the pints A(3, -3) and B(-2,7) is divided by x -axis Also, find the point of division.


The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, -3). The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find the coordinates of another point D such that ABCD is a rhombus.


ΔXYZ ∼ ΔPYR; In ΔXYZ, ∠Y = 60o, XY = 4.5 cm, YZ = 5.1 cm and XYPY =` 4/7` Construct ΔXYZ and ΔPYR.


The measure of the angle between the coordinate axes is


What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4)?

 

Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3).

 

If A (1, 2) B (4, 3) and C (6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of fourth vertex D.

 

A line segment is of length 10 units. If the coordinates of its one end are (2, −3) and the abscissa of the other end is 10, then its ordinate is


The ratio in which (4, 5) divides the join of (2, 3) and (7, 8) is


If the points P (xy) is equidistant from A (5, 1) and B (−1, 5), then


The distance of the point (–6, 8) from x-axis is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×