Advertisements
Advertisements
Question
If the point P(k - 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the value of k.
Advertisements
Solution 1
The given points are P(k - 1, 2), A(3, k) and B(k, 5).
∵ AP = BP
∴ AP2 = BP2
⇒ (k - 1 - 3)2 + (2 - k)2 = (k - 1 - k)2 + (2 - 5)2
⇒ (k - 4)2 + (2 - k)2 = (-1)2 + (-3)2
⇒ k2 - 8k + 16 + 4 + k2 - 4k = 1 + 9
⇒ k2 - 6k + 5 = 0
⇒ (k - 1) (k - 5) = 0
⇒ k = 1 or k = 5
Hence, k = 1 or k = 5
Solution 2
It is given that P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5).
∴ AP = BP
\[\Rightarrow \sqrt{\left[ \left( k - 1 \right) - 3 \right]^2 + \left( 2 - k \right)^2} = \sqrt{\left[ \left( k - 1 \right) - k \right]^2 + \left( 2 - 5 \right)^2} \left( \text{ Distance formula } \right)\]
\[ \Rightarrow \sqrt{\left( k - 4 \right)^2 + \left( 2 - k \right)^2} = \sqrt{\left( - 1 \right)^2 + \left( - 3 \right)^2}\]
Squaring on both sides, we get
\[k^2 - 8k + 16 + 4 - 4k + k^2 = 10\]
\[ \Rightarrow 2 k^2 - 12k + 10 = 0\]
\[ \Rightarrow k^2 - 6k + 5 = 0\]
\[ \Rightarrow \left( k - 1 \right)\left( k - 5 \right) = 0\]
\[\Rightarrow k - 1 = 0 \text{ or k } - 5 = 0\]
k = 1 or k = 5
Thus, the value of k is 1 or 5.
RELATED QUESTIONS
A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is the mid-point of PQ, then find the coordinates of P and Q.
If two opposite vertices of a square are (5, 4) and (1, −6), find the coordinates of its remaining two vertices.
If G be the centroid of a triangle ABC, prove that:
AB2 + BC2 + CA2 = 3 (GA2 + GB2 + GC2)
Determine the ratio in which the straight line x - y - 2 = 0 divides the line segment
joining (3, -1) and (8, 9).
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
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, −3). The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find the coordinates of another point D such that ABCD is a rhombus.
If the points A (2,3), B (4,k ) and C (6,-3) are collinear, find the value of k.
Find the ratio in which the point (−3, k) divides the line-segment joining the points (−5, −4) and (−2, 3). Also find the value of k ?
Find the coordinates of the circumcentre of a triangle whose vertices are (–3, 1), (0, –2) and (1, 3).
Show that A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a
rhombus ABCD.
The co-ordinates of point A and B are 4 and -8 respectively. Find d(A, B).
A point whose abscissa is −3 and ordinate 2 lies in
Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.
If the distance between points (x, 0) and (0, 3) is 5, what are the values of x?
Write the coordinates of the point dividing line segment joining points (2, 3) and (3, 4) internally in the ratio 1 : 5.
What is the distance between the points A (c, 0) and B (0, −c)?
The ratio in which the x-axis divides the segment joining (3, 6) and (12, −3) is
Find the point on the y-axis which is equidistant from the points (5, −2) and (−3, 2).
Signs of the abscissa and ordinate of a point in the second quadrant are respectively.
Abscissa of all the points on the x-axis is ______.
