Advertisements
Advertisements
Question
If the poin A(0,2) is equidistant form the points B (3, p) and C (p ,5) find the value of p. Also, find the length of AB.
Advertisements
Solution
As per the question
AB = AC
`⇒ sqrt((0-3)^2 +(2-p)^2 ) = sqrt((0-p)^2 + (2-5)^2)`
`⇒ sqrt((-3)^2 +(2-p)^2) = sqrt((-p)^2 + (-3)^2)`
Squaring both sides, we get
`(-3)^2 +(2-p)^2 = (-p)^2 +(-3)^2`
`⇒ 9+4+p^2-4p=p^2+9`
`⇒ 4p =4`
⇒ p=1
Now,
`AB = sqrt((0-3)^2 +(2-p)^2)`
`= sqrt((-3)^2 +(2-1)^2))` (∵p=1)
`=sqrt(9+1)`
`= sqrt(10)` units
Hence, p = 1 and AB =`sqrt(10)` units
APPEARS IN
RELATED QUESTIONS
How will you describe the position of a table lamp on your study table to another person?
On which axis do the following points lie?
S(0,5)
The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex.
Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3,5).
Find the points of trisection of the line segment joining the points:
5, −6 and (−7, 5),
Show that the points A (1, 0), B (5, 3), C (2, 7) and D (−2, 4) are the vertices of a parallelogram.
Points P, Q, R and S divide the line segment joining the points A(1,2) and B(6,7) in five equal parts. Find the coordinates of the points P,Q and R
Find the area of quadrilateral PQRS whose vertices are P(-5, -3), Q(-4,-6),R(2, -3) and S(1,2).
Find the area of quadrilateral ABCD whose vertices are A(-3, -1), B(-2,-4) C(4,-1) and D(3,4)
The perpendicular distance of the point P (4, 3) from x-axis is
The area of the triangle formed by the points P (0, 1), Q (0, 5) and R (3, 4) is
ABCD is a parallelogram with vertices \[A ( x_1 , y_1 ), B \left( x_2 , y_2 \right), C ( x_3 , y_3 )\] . Find the coordinates of the fourth vertex D in terms of \[x_1 , x_2 , x_3 , y_1 , y_2 \text{ and } y_3\]
Find the value of k, if the points A(7, −2), B (5, 1) and C (3, 2k) are collinear.
If \[D\left( - \frac{1}{5}, \frac{5}{2} \right), E(7, 3) \text{ and } F\left( \frac{7}{2}, \frac{7}{2} \right)\] are the mid-points of sides of \[∆ ABC\] , find the area of \[∆ ABC\] .
The perimeter of the triangle formed by the points (0, 0), (0, 1) and (0, 1) is
If (x , 2), (−3, −4) and (7, −5) are collinear, then x =
If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then a3 + b3 + c3 =
The points (–5, 2) and (2, –5) lie in the ______.
Point (3, 0) lies in the first quadrant.
