(i) 3y = x + 1 - gradient is 1/3

(ii) x + 3y + 3 = 0 - gradient is -1/3

(iii) x - 3y = 1 - gradient is 1/3

So (i) and (iii) are parallel.

NOTE: it is good practice to calculate all four gradients because if BC // DA, the figure would have been a parallelogram which was not asked.

11.

12. The line joining the points (2, 0) to (0, 4) has gradient of -2 - the rise is 4 and run is 2 and the line slopes down. Hence the tangent to the curve has a gradient of -2.

When x = 1 on the parabola
y = 1 + 2 = 3.

Hence the equation of the tangent to the curve is:

y - 3 = -2(x - 1)

y = -2x + 5

(ii) Gradient is +5.

(iii) Gradient is -5.

(iv) Gradient is -15.

(v) Gradient is +5

Hence lines (i) and (iii) are parallel.

Lines (ii) and (v) are parallel.

16.

(ii)

(iii)

(iv)

(v) The quadrilateral has opposite sides parallel. So it is a parallelogram at least. KL is not perpendicular to LM so the figure is not a rectangle.

The diagonals cut each other at right angles - so the parallelogram can be refined to a rhombus.

See the Quadrilateral Family.

22.

(ii) CT is a horizontal line parallel to RE and passing through y = -3.

Hence CT is y = -3

ET is a vertical line parallel to RC and passing through x = 6.

Hence ET is x = 6.

(i) The sketch gives a hint that the points could be M, P and Q. Calculate the gradients between these points:

Same gradient in each direction and point P links to both M and Q.

∴ M, P and Q are collinear.

(ii) Link the new point to any of the three and calculate the gradient:

(ii)

(iii)

All gradients are the same and each pair is linked.

Hence the four points are collinear.

4y = 8

y = 2

Substitute into Eqn (2):

2x = 0

x = 0

Sub. x = 0 and y = 2 into Eqn (3):

LHS = 2

RHS = 10(0) + 2 = 2 = LHS.

∴ all 3 lines passs through the same point.

Hence they are concurrent. 

Add the 2 eqns:

4x - 12 = 0

x = 3

Substituting into eqn 1:

2y = 3

y = 1.5

POI is (3, 1.5)

(ii) To be concurrent, the new line must pass through the POI:

∴ k(3) + 3(1.5) - 6 = 0

3k = 1.5

k = 0.5.

3x = 3

x = 1

Sub into Eqn (1):

1 + y - 2 = 0

y = 1.

Sub (1, 1) into Eqn (3):

LHS = 1 + 3 - 4 = 0 = RHS

∴ all 3 lines passs through the same point.

Hence they are concurrent. 

(ii)

(iii) Determine the equation of the line perpendicular to each side and passing through the midpoint of each side.

AB:

Midpoint is (1, 0) and gradient is 0 (horizontal. Hence vertical bisector is x = 1.

BC:

Midpoint is (3, 3) and gradient is -3.

Eqn of perpendicular:

AC:

Midpoint is (0, 3) and gradient is 1.5.

Eqn of perpendicular:

(iv) Substitute x = 1 into Eqn BC bisector:

y = 7/3

Substitute x = 1 into Eqn AC bisector:

y = 7/3

All lines meet at the same point so they are concurrent.

36. If concurrent, the two given equations must be solved simultaneously: