Coordinate geometry
Que. 1. Find the distances between the following pairs of points.
1. (2,3), (4,1)
A (2,3) (X1, Y1) B (4,1) (x2,y2)
2. (-5,7) , (-1,3)
P (-5,7) (X1, Y1) Q (-1,3) (x2,y2)
3. (a, b) , (-a, -b)
P (a, b) (X1, Y1) Q (-a, -b) (x2,y2)
Que. 2. Find the distance between the points (0, 0) and (36, 15).
Distance from the origin
Que. 3. Determine whether the points (1, 5), (2, 3), and (-2, -11) are collinear.
A = (1, 5) , B = (2, 3) , C = (-2, -11)
AB + BC = AC
Therefore, these three points are not collinear.
Que. 4. Check whether the points (5, -2), (6, 4), and (7, -2) are the vertices of an isosceles triangle.
A = (5, -2) , B = (6, 4) , C = (7, -2)
AB + BC = AC
Therefore AB = BC
Que. 5. In a classroom, four friends are seated at points A, B, C, and D, as shown in the figure. Champa and Chameli walk into the class and, after observing for a few minutes, Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, determine which of them is correct.
By Distance Formula,
AB = BC = CD = AD; therefore, points A, B, C, and D form a square.
Que. 6. Name the type of quadrilateral formed by the following points (if any) and give reasons for your answer:
1. (-1, -2) , (1, 0) , (-1, 2) , (-3, 0)
By Distance Formula,
AB = BC = CD = AD; therefore, ABCD is a square.
2. (-3, 5) , (3, 1) , (0, 3) , (-1, -4)
By Distance Formula,
Therefore, ABCD is not a quadrilateral.
3. (4, 5) , (7, 6) , (4, 3) , (1, 2)
By Distance Formula,
Hence ABCD is a parallelogram.
Que. 7. Find the point on the x-axis that is equidistant from (2, 5) and (-2, 9).
Point P is equidistant from points A and B; therefore,
AP = BP
Coordinates of point P = (x, 0) = (-7, 0)
Que. 8. Find the value of y for which the distance between the points P(2, 3) and Q(10, y) is 10 units.
Y2 + 6y + 73 = 100
Y2 + 6y + 73 – 100 = 0
Y2 + 6y -27 = 0
Y2 + 9y -3y -27 = 0
Y(y + 9) -3(y + 9) = 0
(y + 9) (y – 3) = 0
Y + 9 = 0 y – 3 = 0
Y = -9 y = 3
Coordinates of point Q = (10, -9) Or
(10, 3)
Que. 9. If Q (0, 1) is equidistant from the points P(5, -3) and R(x, 6), find the value of x. Also, find the distances QR and PR.
41 = x2 + 25
41 – 25 = x2
16 = x2
x2 = 16
Que. 10. Find a relationship between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
On squaring both sides,
X2 + y2 -6x -12y + 45 = x2 + y2 +6x -8y +25
X2 + y2 -6x -12y + 45 -x2 – y2 -6x +8y -25 = 0
-12x + 4y -20 = 0
4 (3x + y -5) = 0
3x + y -5 = 0
