Advertisements
Advertisements
प्रश्न
The distances of point P (x, y) from the points A (1, - 3) and B (- 2, 2) are in the ratio 2: 3.
Show that: 5x2 + 5y2 - 34x + 70y + 58 = 0.
Advertisements
उत्तर
It is given that PA: PB = 2: 3
`"PA"/"PB" = (2)/(3)`
`"PA"^2/"PB"^2 = (4)/(9)`
`((x - 1)^2 + (y + 3)^2)/((x + 2)^2 + (y - 2)^2) = (4)/(9)`
`(x^2 + 1 -2x + y^2 + 9 + 6y)/(x^2 + 4 + 4x + y^2 + 4 - 4y) = (4)/(9)`
9(x2 - 2x + y2 + 10 + 6y) = 4(x2 + 4x + y2 + 8 - 4y)
9x2 - 18x + 9y2 + 90 + 54y = 4x2 + 16x + 4y2 + 32 - 16y
5x2 + 5y2 - 34x + 70y + 58 = 0
Hence, proved.
APPEARS IN
संबंधित प्रश्न
The value of 'a' for which of the following points A(a, 3), B (2, 1) and C(5, a) a collinear. Hence find the equation of the line.
A(–8, 0), B(0, 16) and C(0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5. Show that : PQ = `3/8` BC.
Find the distance of the following points from the origin:
(iii) C (-4,-6)
Using the distance formula, show that the given points are collinear:
(6, 9), (0, 1) and (-6, -7)
Find the distances between the following point.
R(–3a, a), S(a, –2a)
Find the distance between the following pairs of points:
(–3, 6) and (2, –6)
A point P lies on the x-axis and another point Q lies on the y-axis.
Write the ordinate of point P.
Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.
If the distance between the points (4, P) and (1, 0) is 5, then the value of p is ______.
Find the distance between the points O(0, 0) and P(3, 4).
