# Help needed for maths revision questions

Forums Leaving Cert Maths Help needed for maths revision questions

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• #226 Anonymous

:Im having a lot of trouble with the revision questions in the Text and Tests 4 book chapter 3 the Line and we have a test on Monday. I was just wondering if anyone here could help with the qusetions Im stuck on.

5(C)Find the value p if the acute angle between the lines 3x+ty-7=0
2tx-y+5=0 is 45 degrees.

4( Find the coordinates of the orthocentre of the triangle with the vertices (1 (1 -2) (7 1)

6(C) A line containg the point(-2 4) has a slope of m which is not equal to zero. The line intercepts the x-axis at (x1 0) and the y axis at (0 y1)
If x1 + y1=-4m find the slopes of the two lines that satisfy this condition. Hence find the tangent of the acute angle between these two lines.

7(C) Find the equations of the two lines which pass through the point (1 3) if the perpendicular of each line from the origin is 3.

10(c)Find the equations of the lines with the slopes -3/4 and which form a triangle of area 24 square units with the x and y axis.

If any of you could offer me some advice on these questions or show me how to do them that could be great. Thanks for reading this.

• #12526 Anonymous

Im having a lot of trouble with the revision questions in the Text and Tests 4 book chapter 3 the Line and we have a test on Monday. I was just wondering if anyone here could help with the qusetions Im stuck on.

5(C)Find the value p if the acute angle between the lines 3x+ty-7=0
2tx-y+5=0 is 45 degrees.

4( Find the coordinates of the orthocentre of the triangle with the vertices (1 (1 -2) (7 1)

6(C) A line containg the point(-2 4) has a slope of m which is not equal to zero. The line intercepts the x-axis at (x1 0) and the y axis at (0 y1)
If x1 + y1=-4m find the slopes of the two lines that satisfy this condition. Hence find the tangent of the acute angle between these two lines.

7(C) Find the equations of the two lines which pass through the point (1 3) if the perpendicular of each line from the origin is 3.

10(c)Find the equations of the lines with the slopes -3/4 and which form a triangle of area 24 square units with the x and y axis.

If any of you could offer me some advice on these questions or show me how to do them that could be great. Thanks for reading this.

• #12527 Anonymous

QUESTION 5(C) WITHOUT THE QUESTION MARKS ABOVE HOPEFULLY. I THINK THE QUESTION MARKS COME WHEN THERE IS A GAP BETWEEN THE MINUS AND THE DIGIT

Call line 3x + ty?7 = 0 line K. Put 3x + ty?7 = 0 in the form y = mx + c. K would then read ty = -3x + 7. Divid

• #12528 Anonymous

if you go to http://www.examinations.ie and click on examinations material archive tick the accepting terms nd conditions box click 2005 in the drop down box and then marking schemes…if after that you can find the 2005 leaving cert higher maths paper 2 questi

• #12529 Anonymous

hi

ive done 4( the one about the orthocentre. i cant put the diagram on here so if you want it on word you can email me at @yahoo.co.uk">pathwayanalysis@yahoo.co.uk

best of luck mon

• #12530 Anonymous

we dont have that book. could you give us the answers….hopefully they are in the back of your book and we will know what direction we are heading in. thanks

• #12531 Anonymous

7(c) one of the lines is defo y = 3 cos it would be parallel to the x-axis and at (03) it would be 3 from da origin nd it goes thro (13)

• #12532 Anonymous

Question 4(b)

We have a(18) b(1-2) c(71)

Slope ac = -7/6

So slope of perpendicular line to ac from opposite vertice b to point d on ac is 6/7

Eqn of line bd is

Y ?y1 = m(x-x1)

Y ?(-2) = (6/7)(x- 1)

Y + 2 = 6x/7 ? 6/7

Multiply ac

• #12533 Anonymous

see all the question marks in the last comment. they were entered as minuses but dunno why they turned into question marks

• #12534 Anonymous

Thanks everyone for all the help and support. The answers to the questions from the back of the book are 5c. t= +- 1+- 1.5
4b. (4.51)
6c. m=-12; tangent=3
7c. y=3;3x+4y-15=0
10c.3x+4y-24=0; 3x+4y+24=0

• #12535 Anonymous

in 5(c) is it find the value t and not p? maybe im really thick but i cant see a p anywhere. thanks

• #12536 Anonymous

QUESTION 5(C). Call line 3x + ty ?7 = 0 like K. Put 3x + ty ?7 = 0 in the form y = mx + c. K would then read ty = -3x + 7. Dividing by t gives y = (-3/t)x + (7/t). So the slope of K is (-3/t). Call this slope m1. Call line 2tx ? y + 5 = 0 line L. Puttin

• #12537 Anonymous

DAMN STILL HAVE QUESTION MARKS. ANYWAY YOU WILL HAVE TO SUBSTITUTE MINUSES FOR THE QUESTION MARKS. I WILL TRY TO EMAIL THIS VERSION COS THE ATTACHMENTS ARE NOT GETTING THROUGH ND HOPEFULLY THERE WONT BE QUESTION MARKS ON THE EMAIL IF IT GETS THROUGH.

• #12538 Anonymous

Oh Im really sorry it was t and not p.

• #12539 Anonymous

I made a mistake in Q6c It should have been x1 + y1=-4 not x1 + y1=-4m. Thanks everyone for all your help.