Preliminary Maths Questions (updated)

Hi, I need help with a few questions, any feedback would be greatly appreciated.

1. Determine the equation of the line through M(-1, 2) if M is the mid-point of the intercepts on the x-axis and y-axis.

2. Solve the inequation x² - 4>0 and hence state the domain of 5/√x² -4

3. set up a pair of simutaneous equations and hence solve the problem :
two people are 16km apart on a straight road. they start walking at the same time, If they walk towards each other they will meet in 2hrs but if they walk in the same direction (so that the distance between them is decreasing, they will meet in 8hrs. find their walking speeds.

thanks in advance for any help you can give.
:)

4. solve the following for x and y given both are rational numbers
1½ + √x+2y = 3x+y+2/3√3

for this question i put the rhs in to y=mx+b form and got y= -3x - 2/3√3 and then subbed that into the left hand side. so i got 1½ +√x+2(-3x- 2/3√3)
then i expanded the bracket and got 1½ + √x-6x - 4/3√3 but from there i got lost, i couldnt find a way to get rid of the surds to make them rational numbers.

5. determine the length of the chord cut off from the line 2y -x +8 =0 by the circle (x+2)² + (y+3)² = 4.

i got that (-2, -3) is the centre of the circle and that its radius is two. i put the line in y= mx+b form and sub some values in for x and then my line ended up no where near my circle. is there an easier way to do this?


thanks again.

1. Midpoint of two points (x1, y1) and (x2, y2) is ([x1 - x2]/2, [y1 - y2]/2)

Your points are (x1, 0) x-intercept and (0, y2) y-intercept

So -1 = [x1 - 0]/2
2 = [0 - y2]/2

Solve for x1 and y2
Gradient = Change in y / Change in x = m

Put into y = mx + c with c = y2

2. Factorise x² - 4 so you can sketch the graph of y = x² - 4

The range of x values for which the graph is above the y-axis is the answer for the inequality

The square root of x² - 4 is only possible if x² - 4 is positive so the domain is the same as the answer to the inequality

3. Let speed of person 1 = v1 and speed of person 2 = v2

Speed = Distance/Time so Distance = Speed x Time

Walking towards each other Person 1 travels 2v1 and Person 2 travels 2v2
They both walk a combined distance of 16km so 2v1 + 2v2 = 16

Walking same direction Person 1 towards Person 2
Person 1 travels 8v1 andPerson 2 travels 8v2
Person 1 walks 16km more than Person 2 so 8v1 - 8v2 = 16

Once you have these simultaneous equations solve as normal

5. Rewrite straight line as x = 2y + 8

Substitute for x in the equation of the circle

(2y + 8 + 2)² + (y + 3)² = 4

Expand the brackets and solve for y. It is a quadratic so there will be 2 solutions which will correspond to the y coordinates where the line crosses the circles

Put y coordinates into x = 2y + 8 to get the x coordinates

You will then have the two coordinates where the line crosses the circle (x1, y1) and (x2, y2)

Use Pythagoras to find length of chord √[(x1 - x2)² + (y1 - y2)²]

4. Can't treat RHS as separate equation

Collect like terms

-3x + √(x) + y = 2/3√3 - 1½

Simplify RHS

-3x + √(x) + y = 2√3/9 - 3/2 = 4√3/18 - 27/18 = (4√3 - 27)/18

Multiply both sides by 18

-54x + 18√(x) + 18y = 4√3 - 27

Make 18√(x) = 4√3 because x and y are rational so the only way of getting irrationals on the LHS and so -54x + 18y = -27

√324 √(x) = √16 √3

Use rules of surds

√(324x) = √48

324x = 48

x = 48/324 = 24/162 = 12/81 = 4/27

Substitute x value into -54x + 18y = -27

-216/27 + 18y = -27

Same as -24/27 + 2y = -3

Solves to 2y = -57/27 giving y = -57/54 = -19/18

Check my working as I'm concentrating on another task as I do this but the reasoning is sound