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January 1, 1970 at 5:29 am #226Anonymous
: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+ty7=0
2txy+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 xaxis 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.

January 1, 1970 at 5:29 am #12526Anonymous
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+ty7=0
2txy+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 xaxis 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.

January 1, 1970 at 5:29 am #12527Anonymous
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

January 1, 1970 at 5:29 am #12528Anonymous
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

January 1, 1970 at 5:29 am #12529

January 1, 1970 at 5:29 am #12530Anonymous
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

January 1, 1970 at 5:29 am #12531Anonymous
7(c) one of the lines is defo y = 3 cos it would be parallel to the xaxis and at (03) it would be 3 from da origin nd it goes thro (13)

January 1, 1970 at 5:29 am #12532Anonymous
Question 4(b)
We have a(18) b(12) 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(xx1)
Y ?(2) = (6/7)(x 1)
Y + 2 = 6x/7 ? 6/7
Multiply ac

January 1, 1970 at 5:29 am #12533Anonymous
see all the question marks in the last comment. they were entered as minuses but dunno why they turned into question marks

January 1, 1970 at 5:29 am #12534Anonymous
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+4y15=0
10c.3x+4y24=0; 3x+4y+24=0 
January 1, 1970 at 5:29 am #12535Anonymous
in 5(c) is it find the value t and not p? maybe im really thick but i cant see a p anywhere. thanks

January 1, 1970 at 5:29 am #12536Anonymous
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

January 1, 1970 at 5:29 am #12537Anonymous
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.

January 1, 1970 at 5:29 am #12538Anonymous
Oh Im really sorry it was t and not p.

January 1, 1970 at 5:29 am #12539Anonymous
I made a mistake in Q6c It should have been x1 + y1=4 not x1 + y1=4m. Thanks everyone for all your help.


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