Advertisements
Advertisements
प्रश्न
Find the value of k, if the points A(7, −2), B (5, 1) and C (3, 2k) are collinear.
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(7, −2), B(5, 1) and C(3, 2k). It is also said that they are collinear and hence the area enclosed by them should be 0.
\[∆ = \frac{1}{2}\left| \left( 7 \times 1 + 5 \times 2k + 3 \times - 2 \right) - \left( 5 \times - 2 + 3 \times 1 + 7 \times 2k \right) \right|\]
\[ 0 = \frac{1}{2}\left| \left( 7 + 10k - 6 \right) - \left( - 10 + 3 + 14k \right) \right|\]
\[ 0 = \frac{1}{2}\left| - 4k + 8 \right|\]
\[ 0 = - 4k + 8\]
\[ k = 2\]
Hence the value of ‘k’ for which the given points are collinear is k = 2 .
APPEARS IN
संबंधित प्रश्न
(Street Plan): A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.
All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines.
There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
- how many cross - streets can be referred to as (4, 3).
- how many cross - streets can be referred to as (3, 4).
Which point on the y-axis is equidistant from (2, 3) and (−4, 1)?
Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6,2) are the vertices of a square.
If three consecutive vertices of a parallelogram are (1, -2), (3, 6) and (5, 10), find its fourth vertex.
Find the coordinates of the points which divide the line segment joining the points (-4, 0) and (0, 6) in four equal parts.
Determine the ratio in which the point (-6, a) divides the join of A (-3, 1) and B (-8, 9). Also, find the value of a.
Find the co-ordinates of the point which divides the join of A(-5, 11) and B(4,-7) in the ratio 7 : 2
Point A lies on the line segment PQ joining P(6, -6) and Q(-4, -1) in such a way that `(PA)/( PQ)=2/5` . If that point A also lies on the line 3x + k( y + 1 ) = 0, find the value of k.
If (2, p) is the midpoint of the line segment joining the points A(6, -5) and B(-2,11) find the value of p.
The abscissa and ordinate of the origin are
The area of the triangle formed by the points A(2,0) B(6,0) and C(4,6) is
ABCD is a parallelogram with vertices \[A ( x_1 , y_1 ), B \left( x_2 , y_2 \right), C ( x_3 , y_3 )\] . Find the coordinates of the fourth vertex D in terms of \[x_1 , x_2 , x_3 , y_1 , y_2 \text{ and } y_3\]
If three points (x1, y1) (x2, y2), (x3, y3) lie on the same line, prove that \[\frac{y_2 - y_3}{x_2 x_3} + \frac{y_3 - y_1}{x_3 x_1} + \frac{y_1 - y_2}{x_1 x_2} = 0\]
If the points A (1,2) , O (0,0) and C (a,b) are collinear , then find a : b.
The distance between the points (cos θ, 0) and (sin θ − cos θ) is
If points (a, 0), (0, b) and (1, 1) are collinear, then \[\frac{1}{a} + \frac{1}{b} =\]
If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then a3 + b3 + c3 =
If the coordinates of the two points are P(–2, 3) and Q(–3, 5), then (abscissa of P) – (abscissa of Q) is ______.
Find the coordinates of the point which lies on x and y axes both.
The distance of the point (–4, 3) from y-axis is ______.