Advertisements
Advertisements
Question
If the point P (m, 3) lies on the line segment joining the points \[A\left( - \frac{2}{5}, 6 \right)\] and B (2, 8), find the value of m.
Advertisements
Solution
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.
It is said that the point P(m,3) lies on the line segment joining the points A`(-2/5,6)` and B(2,8). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.
\[∆ = \frac{1}{2}\left| \left( - \frac{2}{5} \times 3 + m \times 8 + 2 \times 6 \right) - \left( m \times 6 + 2 \times 3 + \left( - \frac{2}{5} \right) \times 8 \right) \right|\]
\[ 0 = \frac{1}{2}\left| \left( - \frac{6}{5} + 8m + 12 \right) - \left( 6m + 6 - \frac{16}{5} \right) \right|\]
\[ 0 = \frac{1}{2}\left| 2m + 8 \right|\]
\[ 0 = 2m + 8\]
\[ m = - 4\]
Hence the value of ‘m’ for which the given condition is satisfied is m = - 4 .
APPEARS IN
RELATED QUESTIONS
In Fig. 14.36, a right triangle BOA is given C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.
We have a right angled triangle,`triangle BOA` right angled at O. Co-ordinates are B (0,2b); A (2a, 0) and C (0, 0).
Name the quadrilateral formed, if any, by the following points, and given reasons for your answers:
A(4, 5) B(7, 6), C (4, 3), D(1, 2)
Find the area of quadrilateral PQRS whose vertices are P(-5, -3), Q(-4,-6),R(2, -3) and S(1,2).
If the points A (2,3), B (4,k ) and C (6,-3) are collinear, find the value of k.
Find the coordinates of the centre of the circle passing through the points P(6, –6), Q(3, –7) and R (3, 3).
The ordinate of any point on x-axis is
If the points P, Q(x, 7), R, S(6, y) in this order divide the line segment joining A(2, p) and B(7, 10) in 5 equal parts, find x, y and p.
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 mid-point of the segment joining A (x, y + 1) and B (x + 1, y + 2) is C \[\left( \frac{3}{2}, \frac{5}{2} \right)\] , find x, y.
Write the coordinates the reflections of points (3, 5) in X and Y -axes.
Find the value of a so that the point (3, a) lies on the line represented by 2x − 3y + 5 = 0
If the centroid of a triangle is (1, 4) and two of its vertices are (4, −3) and (−9, 7), then the area of the triangle is
If P is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point Q on OY such that OP = OQ, are
If (−2, 1) is the centroid of the triangle having its vertices at (x , 0) (5, −2), (−8, y), then x, y satisfy the relation
f the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (−2, 5), then the coordinates of the other end of the diameter are
The point R divides the line segment AB, where A(−4, 0) and B(0, 6) such that AR=34AB.">AR = `3/4`AB. Find the coordinates of R.
The distance of the point P(2, 3) from the x-axis is ______.
Abscissa of a point is positive in ______.
The distance of the point (3, 5) from x-axis (in units) is ______.
