Maths, Addmaths and Statistics: Post your doubts here!

[abcde]

Thanks!

but I have a problem how did you get the gradient of first line and second line?

Rearrange the equations into the form y = mx + c, where m is the gradient of a straight line.
After rearranging, you get:
1. y = 2/k x - 5/k (first line)
2. y = (k+1)/6 x + 3/6 (second line)
So the gradients are the values where 'm' is found: 2/k and (k+1)/6, respectively.

 

[Waleed007]

Got it Thanks

 

[ailg1996]

can sum1 tell me how to draw a tangent properly i always get the wrong gradient???/

 

[abcde]

can sum1 tell me how to draw a tangent properly i always get the wrong gradient???/

AoA!
Finding the gradient of a curve at a given point using graphical means will not bear an accurate answer. Fortunately, CIE always specifies a range in which your answer must lie so you don't need to worry about it....much . Ways to improve your gradient calculation:
-> Ensure that the tangent touches the curve at a SINGLE point.
-> Draw a tangent that covers at least half of your graph. In fact, the larger the triangle for calculating the gradient, the more accurate the answer. :wink:
-> Use dotted lines for drawing the other sides of the triangle (the tangent is the hypotenuse) and mention the coordinates of the vertices on your graph. Doing so will reduce chances of error while calculating the gradient.

 

[Anon]

can sum1 tell me how to draw a tangent properly i always get the wrong gradient???/

Do as abcde said.

But also, of you are a Add.Math Student, you can find derivative, to know what the actual gradient is... make sure your gradient is somewhere near it...

If you are not, Learn how to calculate derivative in the calculator. You just need to key in the equation, and the x coordinate, and you get the gradient...

This will help you...

 

[Waleed007]

Here's my Question regarding coordinate geometry:
PQRS is a square whose vertices are p(1,6),Q(2,1),R(7,2) and S(h,k)
(a) Fine the values of h and k
(b) Calculate the area of PQRS.

 

[scouserlfc]

First draw a normal square and then ull get the whole picture !
Easy first find midpoint of PR and then put these in midpoint formula for QS and find !
PR midpoint is 4,4

now use these to find the value of h and k by means of QS !
h=6 and k=7

now find length of PQ or any side and then square and u will see an answer !

 

[Waleed007]

Dude Second part i didn't get first u find the midpoint of PR. and then put these in midpoint formula for QS right! but the Coordinates of Q is already given?

 

[talhajohar]

Here's my Question regarding coordinate geometry:
PQRS is a square whose vertices are p(1,6),Q(2,1),R(7,2) and S(h,k)
(a) Fine the values of h and k
(b) Calculate the area of PQRS.

 

[Anon]

@Waleed:

Find Mid point of PR...

Now QS will have the same midpoint as PR.

so by putting the formulae for midpoint, you can find value of h and k

 

[Waleed007]

Here another Question:
P is the point (2,3) and Q is the point (9,5).
(a):find the coordinates of the point where the line PQ intersects the x-axis
(b):The line y=5 is the line of symmetry of triangle PQR.Find the coordinates of R.
(c):Calculate the area of Triangle PQR.

 

[SalmanPakRocks]

Here another Question:
P is the point (2,3) and Q is the point (9,5).
(a):find the coordinates of the point where the line PQ intersects the x-axis
(b):The line y=5 is the line of symmetry of triangle PQR.Find the coordinates of R.
(c):Calculate the area of Triangle PQR.

a) y=mx+c
first find m and c, for it input value of x and y
m= y2-y1/x2-x1
=2/7
for c
3=2/7(2) + c
c = 2.4

ok now input y = 0 as the point where the line will cross x axis y will be 0
0=2/7(x) + 2.4
2x/7 + 2.4
2x/7 = -2.4
2x = -2.4x7
x= -8.4
so points will be (-8.4,0)

 

[SalmanPakRocks]

for part b i draw a figure, then made line y=5
answer is (2,7)

 

[Waleed007]

Cant we take y-y1=m(x-x1) instead of y=mx+c

 

[SalmanPakRocks]

Area = 1/2 * b * h
1/2 * 4 * 7
=14 cm^2

 

[SalmanPakRocks]

Cant we take y-y1=m(x-x1) instead of y=mx+c

That is for Add-Maths I guess. In maths the general form of equation of a line is y=mx+c

 

[Waleed007]

Area = 1/2 * b * h
1/2 * 4 * 7
=14 cm^2

how did you take that Area?

 

[SalmanPakRocks]

Make a rough sketch and it will be lot easier for u to understand. just count the boxes and u'll find the height and the base

 

[scouserlfc]

Always Waleed do this make a rough sketch this helps u understand a lot better ! Believe me !

 

[abcde]

Cant we take y-y1=m(x-x1) instead of y=mx+c

That is for Add-Maths I guess. In maths the general form of equation of a line is y=mx+c

The gradient-intercept form is y = mx + c. You can use y-y1=m(x-x1) anywhere unless otherwise specified.