Advertisements
Advertisements
Question
Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.
Advertisements
Solution
AB = 5
AB2 = 25
(0 - 3)2 + (y - 1 - 1)2 = 25
9 + y2 + 4 - 4y = 25
y2 - 4y - 12 = 0
y2 - 6y + 2y - 12 = 0
y(y - 6) + 2(y - 6) = 0
(y - 6) (y + 2) = 0
y = 6, -2
APPEARS IN
RELATED QUESTIONS
If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.
Distance of point (−3, 4) from the origin is ______.
Prove that the points (5 , 3) , (1 , 2), (2 , -2) and (6 ,-1) are the vertices of a square.
Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
Point P (2, -7) is the center of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of: AT
Find distance between points O(0, 0) and B(– 5, 12)
Find distance between point Q(3, – 7) and point R(3, 3)
Solution: Suppose Q(x1, y1) and point R(x2, y2)
x1 = 3, y1 = – 7 and x2 = 3, y2 = 3
Using distance formula,
d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `sqrt(square - 100)`
∴ d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `square`
Find distance between point A(–1, 1) and point B(5, –7):
Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = – 7
Using distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ d(A, B) = `sqrt(square +[(-7) + square]^2`
∴ d(A, B) = `sqrt(square)`
∴ d(A, B) = `square`
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 ______.
Find the points on the x-axis which are at a distance of `2sqrt(5)` from the point (7, – 4). How many such points are there?
