Coordinate Geometry — Exercise 7.1

Apply the distance formula separately to each pair.

\[\begin{aligned}(i);&\sqrt{(4-2)^2+(1-3)^2}=\sqrt8=2\sqrt2\\(ii);&\sqrt{(-1+5)^2+(3-7)^2}=\sqrt{32}=4\sqrt2\\(iii);&\sqrt{(-2a)^2+(-2b)^2}=2\sqrt{a^2+b^2}\end{aligned}\]

Distance=√(36²+15²)=√(1296+225)=√1521=39. The model uses kilometres on both axes.

Let these be A,B,C. AB²=1²+(−2)²=5, BC²=(−4)²+(−14)²=212 and AC²=(−3)²+(−16)²=265. Alternatively slopes AB=−2 and AC=16/3 differ. No common straight line passes through all three.

AB²=1²+6²=37 and BC²=1²+(−6)²=37, whereas AC=2. The points are not collinear: B is above the horizontal line AC.

Read A(3,4), B(6,7), C(9,4), D(6,1). Every side has squared length 3²+3²=18. Both diagonals have length 6. A quadrilateral with four equal sides and equal diagonals is a square (adjacent vectors (3,3) and (3,−3) are also perpendicular).

(i) Consecutive squared sides are all 8; diagonals are both 4, so the figure is a square. (ii) A(−3,5),B(3,1),C(0,3) are collinear: C is the midpoint of AB. Three collinear vertices cannot form a non-degenerate quadrilateral. (iii) AB²=CD²=10, BC²=DA²=18. Diagonal midpoints both equal (4,4), so the quadrilateral is a parallelogram. Its unequal adjacent sides exclude a rhombus; its unequal diagonals (lengths 2 and √52) exclude a rectangle.

Write the point as (x,0). Equate squared distances: (x−2)²+25=(x+2)²+81. Expand and cancel x²+4 to obtain −4x+25=4x+81, hence −8x=56.

Square the distance relation: 10²=(10−2)²+(y+3)². Thus (y+3)²=36; y+3=±6.

QP²=5²+(−4)²=41. QR²=x²+5². Hence x²+25=41, so x=±4 and QR=√41. For x=4, PR²=(4−5)²+(6+3)²=82. For x=−4, PR²=(−4−5)²+9²=162.

Equate squared distances: (x−3)²+(y−6)²=(x+3)²+(y−4)². Expanding and cancelling yields −12x−4y+20=0. Divide by −4.