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14. Co-ordinate Geometry
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Q2 of 149 Page 14

Find the value of a when the distance between the points (3, a) and (4, 1) is .

Distance =



= 10



a2 - 2a - 8 = 0


a2 - 4a + 2a – 8 = 0


a(a - 4) + 2(a - 4) = 0


(a - 4) (a + 2) = 0


a = 4 and a = -2


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3

The base PQ of two equilateral triangles PQR and PQR' with side 2a lies along y-axis such that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R' of the triangles.

1

Find the distance between the following pair of points:

(i) (-6,7) and (-1, -5)

(ii) (a + b, b + c) and (a - b, c - b)

(iii) (a sin a, - b cos a) and (-a cos a, b sin a)

(iv) (a, 0) and (0, b)

3

If the points (2, 1) and (1,-2) are equidistant from the point (x, y), show that x + 3y = 0.

4

Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.

Questions · 149
14. Co-ordinate Geometry
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