Q5. Find the mid-point of the segment joining (3, −7) and (−5, 1).
((3 − 5)/2, (−7 + 1)/2) = (−1, −3).
Q6. For what value of k are the points (1, −1), (2, 1) and (k, 5) collinear?
1(1 − 5) + 2(5 + 1) + k(−1 − 1) = −4 + 12 − 2k = 8 − 2k = 0, so k = 4.
Q7. Find the point on the y-axis which is equidistant from (−5, −2) and (3, 2).
Take (0, y): 25 + (y + 2)2 = 9 + (y − 2)2 gives 25 + 4y + 4 = 9 − 4y + 4, so 8y = −16 and y = −2. The point is (0, −2), at distance 5 from both.
Q8. In what ratio does the y-axis divide the segment joining (5, −6) and (−1, −4)? Find the point of division.
Take the ratio k : 1. The x-coordinate (−k + 5)/(k + 1) = 0 gives k = 5, so the ratio is 5 : 1. The point is (0, (5 × (−4) + (−6))/6) = (0, −13/3).
Q9. Show that (1, 7), (4, 2), (−1, −1) and (−4, 4), taken in order, are the vertices of a square.
Each side is √34: for example, (4 − 1)2 + (2 − 7)2 = 9 + 25 = 34. Both diagonals are √68: (−1 − 1)2 + (−1 − 7)2 = 4 + 64 and (−4 − 4)2 + (4 − 2)2 = 64 + 4. Four equal sides and equal diagonals make a square.
Q10. The points (1, 2), (4, y), (x, 6) and (3, 5), taken in order, are the vertices of a parallelogram. Find x and y.
The diagonals bisect each other, so the mid-point of (1, 2) and (x, 6) equals the mid-point of (4, y) and (3, 5): (1 + x)/2 = 7/2 and 8/2 = (y + 5)/2. So x = 6 and y = 3.
Q11. Find the area of the triangle formed by the mid-points of the sides of the triangle with vertices (0, −1), (2, 1) and (0, 3). Compare it with the area of the whole triangle.
The mid-points are (1, 0), (1, 2) and (0, 1). Their triangle has area ½|1(2 − 1) + 1(1 − 0) + 0(0 − 2)| = 1. The whole triangle has area ½|0(1 − 3) + 2(3 + 1) + 0(−1 − 1)| = 4. The ratio is 1 : 4.
Q12. Show that the area of the triangle with vertices (t, t − 2), (t + 2, t + 2) and (t + 3, t) does not depend on t.
The bracket is t(t + 2 − t) + (t + 2)(t − t + 2) + (t + 3)(t − 2 − t − 2)
= 2t + 2t + 4 − 4t − 12 = −8,
so the area is 4 square units for every t.
Q13. Find the point dividing the segment joining (1, 2) and (4, 8) externally in the ratio 3 : 1, and check your answer with distances.
((3 × 4 − 1 × 1)/2, (3 × 8 − 1 × 2)/2) = (11/2, 11). Its distances from (1, 2) and (4, 8) are √(81/4 + 81) = (9/2)√5 and √(9/4 + 9) = (3/2)√5, in the ratio 3 : 1.