Advertisements
Advertisements
प्रश्न
Find the distance between the following pairs of points:
(–3, 6) and (2, –6)
Advertisements
उत्तर
(–3, 6) and (2, –6)
x1 = –3, y1 = 6, x2 = 2, y2 = –6
AB = `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
= `sqrt((2 + 3)^2 + (-6 -6)^2)`
= `sqrt((5)^2 + (-12)^2)`
= `sqrt(25 + 144)`
= `sqrt(169)`
= 13
APPEARS IN
संबंधित प्रश्न
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, –5) and R(–3, 6), find the coordinates of P.
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Prove that the points A(1, 7), B(4, 2), C(−1, −1), and D(−4, 4) are the vertices of a square.
Find the distance between the points:
P(a sin α, a cos α) and Q(a cos α, – a sin α)
Show that the points A(1, 2), B(1, 6), C(1 + 2`sqrt3`, 4) are vertices of an equilateral triangle.
ABCD is a square . If the coordinates of A and C are (5 , 4) and (-1 , 6) ; find the coordinates of B and D.
Points A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.
Find distance between points O(0, 0) and B(–5, 12).
∆ABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to ∆DEF with vertices D(–4, 0), E(4, 0) and F(0, 4).
Find distance between points P(– 5, – 7) and Q(0, 3).
By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(125)`
= `5sqrt(5)`
