Advertisements
Advertisements
प्रश्न
Using the distance formula, show that the given points are collinear:
(6, 9), (0, 1) and (-6, -7)
उत्तर
Let A( 6,9) ,B( 0,1 ) and C (-6, -7) be the give points. Then
`AB = sqrt((0-6)^2 +(1-9)^2 )= sqrt((-6)^2 +(-8)^2) = sqrt(36 + 64)= sqrt(100)`=10 units
`BC=sqrt((-6-0)^2+(-7-1)^2) = sqrt((-6)^2+(-8)^2)= sqrt(36 + 64)= sqrt(100) `=10 units
`AC = sqrt((-6-6)^2 +(-7-9)^2 )= sqrt((-12)^2 +(16)^2) = sqrt(144 + 256 )= sqrt(400)`= 20 units
∴ AB + BC = (10+10) units - 20 units = AC
Hence, the given points are collinear
APPEARS IN
संबंधित प्रश्न
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
If the point P(x, y ) is equidistant from the points A(5, 1) and B (1, 5), prove that x = y.
AB and AC are the two chords of a circle whose radius is r. If p and q are
the distance of chord AB and CD, from the centre respectively and if
AB = 2AC then proove that 4q2 = p2 + 3r2.
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 .
Show that P(– 2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle
Show that the point (0, 9) is equidistant from the points (– 4, 1) and (4, 1)
Show that the points (0, –1), (8, 3), (6, 7) and (– 2, 3) are vertices of a rectangle.
Show that A(1, 2), (1, 6), C(1 + 2 `sqrt(3)`, 4) are vertices of a equilateral triangle
A circle drawn with origin as the centre passes through `(13/2, 0)`. The point which does not lie in the interior of the circle is ______.
The distance between the points A(0, 6) and B(0, –2) is ______.