Advertisements
Advertisements
Question
Find the distances between the following point.
P(–6, –3), Q(–1, 9)
Advertisements
Solution
P(–6, –3), Q(–1, 9)
\[PQ = \sqrt{\left( - 6 - \left( - 1 \right) \right)^2 + \left( - 3 - 9 \right)^2}\]
\[ = \sqrt{25 + 144}\]
\[ = \sqrt{169}\]
\[ = 13\]
APPEARS IN
RELATED QUESTIONS
If a ≠ b ≠ 0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.
Find the value of a when the distance between the points (3, a) and (4, 1) is `sqrt10`
Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.
Find the distance between the following pair of points.
L(5, –8), M(–7, –3)
Find the distance of a point (13 , -9) from another point on the line y = 0 whose abscissa is 1.
Show that the points (2, 0), (–2, 0), and (0, 2) are the vertices of a triangle. Also, a state with the reason for the type of triangle.
Find the distance between the origin and the point:
(8, −15)
The distance between the points (3, 1) and (0, x) is 5. Find x.
A point A is at a distance of `sqrt(10)` unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.
Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.
Point P (2, -7) is the centre of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of AB.
Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
Find distance between point A(7, 5) and B(2, 5).
The point which lies on the perpendicular bisector of the line segment joining the points A(–2, –5) and B(2, 5) is ______.
Point P(0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A(–1, 1) and B(3, 3).
The points A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0) in that order form a rectangle.
The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, – 9) and has diameter `10sqrt(2)` units.
Find the distance between the points O(0, 0) and P(3, 4).
Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.
