RD Chapter 14 Co ordinate Geometry Ex 14.2 Solutions
Question - 21 : - Find a point which is equidistant from the point A (-5, 4) and B (-1, 6). How many such points are there? [NCERT Exemplar]
Answer - 21 : -
Let P (h, k) be the point which is equidistant from the points A (-5, 4) and B (-1, 6).
PA = PB
So, the mid-point of AB satisfy the Eq. (i).
Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the point A and B.
Replacing h, k, by x, y in above equation, we have 2x + y + 1 = 0
Question - 22 : - The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units. [NCERT Exemplar]
Answer - 22 : -
By given condition,
Distance between the centre C (2a, a-1) and the point P (11, -9), which lie on the circle = Radius of circle
Question - 23 : - Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distance covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in kilometers. [NCERT Exemplar]
Answer - 23 : -
Question - 24 : - Find the value of k, if the point P (0, 2) is equidistant from (3, k) and (k, 5).
Answer - 24 : -
Let P (0, 2) is equidistant from A (3, k) and B (k, 5)
PA = PB
=> PA² = PB²
Question - 25 : - If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the
(i) interior,
(ii) exterior of the triangle. [NCERT Exemplar]
Answer - 25 : -
Let the third vertex of an equilateral triangle be (x, y).
Let A (-4, 3), B (4,3) and C (x, y).
We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal.
But given that, the origin lies in the interior of the ∆ABC and the x-coordinate of third vertex is zero.
Then, y-coordinate of third vertex should be negative.
Hence, the require coordinate of third vertex,
C = (0, 3 – 4√3). [C ≠ (0, 3 + 4√3)]
Question - 26 : - Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.
Answer - 26 : - Let the co-ordinates of the vertices A, B, C and D of a rhombus are A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4)
Question - 27 : - Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.
Answer - 27 : - Let ABC is a triangle whose vertices are A (3, 0), B (-1, -6) and C (4, -1)
Question - 28 : - Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4). [CBSE 2005]
Answer - 28 : -
The required point is on x-axis
Its ordinate will be O
Let the co-ordinates of the required point P (x, 0)
Let the point P is equidistant from the points A (7, 6) and B (-3, 4)
Question - 29 : - (i) Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square. [CBSE 2004]
(ii) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD. [CBSE 2013]
(iii) Name the type of triangle PQR formed by the point P(√2 , √2), Q(- √2, – √2) and R (-√6 , √6 ). [NCERT Exemplar]
Answer - 29 : -
Question - 30 : - Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3). [CBSE 2004]
Answer - 30 : -
The point P lies on x-axis
The ordinates of P will be 0 Let the point P be (x, 0)
Let P is equidistant from A (-2, 5) and B (2, -3)