PLease help me...?

Question

Question: Given that the points (-2,8), (1,1) and (3,k) lie on a straight line, calculate the value of k.


Note: Please help me. I know that the 1st number of each are the x-values so k is a y-value. How do I solve this but also how do I clearly show the working?

Could you please take me through the steps of finding out what k is (the value). Thankyou so much for your help.

Question 2: I need to express the equation x/3 + y/6 = 1 in the form y=mx+c

What is the answer? What I did, was I multiplied the x/3 by two so that the denominator of the fraction was 6 like y/6. I then multiplied the 2x/6, y/6 and 1 by 6 so that the two (y and 2x) on the left hand side of the equals sign were whole numbers. I subtracted 2 x from the left hand side so it was -2x on the right hand side. This worked out to: y=-2x+6 Is this correct?

Thankyou!

Oh..also, when I am asked for the gradient and y-intercept of an equation in the form, y=mx+c if the gradient and y-intercept are fractions, do I leave them as fractions or do I convert them into decimals correct to 2dp?

to answer the question?

Answer 1

#1)
Let's use the two known points to derive an equation:
slope (gradient) is found by:
m = (y1-y2)/(x1-x2)
= (8-1)/(-2-1)
= 7/-3
= -7/3

The slope between any of the points will be the same because they're all on the same line. So now, calculate the slope using one of the known points and the unknown. (1, 1) and (3, k)

m = (k-1)/(3-1)
= (k-1)/2

The slopes are equal, so
-7/3 = (k-1)/2
-14/3 = k-1
-14/3 + 1 = k
-14/3 + 3/3 = k
-11/3 = k
------------------
#2
x/3 + y/6 = 1
Let's first get rid of the fractions by multiplying everything by 6.

2x + y = 6
y = -2x + 6

Yes, you got that right!!!

Now that it's in that form you can find just about anything they want.

the general form is y=mx+b. the m is the gradient (slope). b is the y-intercept.

So, m = -2, b = 6.

If you need to find the x-intercept, that's where y=0. So,
0 = -2x + 6
2x = 6
x = 3

Answer 2

1. (-2,8), (1,1) and (3,k)
equation for straight line: y=mx +c
substituing the values:
8= -2m + c eq i
1= m + c eq ii subtracting, u get
---------------
7 = -3m
or, m = -7/3 (gradient)
substituting this valueof m in eq i
8= -2(-7/3) + c
8= 14/3 + c
c = 8 - 14/3 = (24-14)/3 =10/3

now, the equation can be re-written as
y = -7/3 x + 10/3

putting 3rd set of co-ordinates (3,k)
k= -(7/3)3 + 10/3
k= -21/3 +10/3
k= (-21+10)/3 = -11/3 Ans

2. x/3 + y/6 = 1
multiplying both sides (RHS & LHS by 6) we get

6(x/3) +6(y/6) = 1(6)
or, 2x + y = 6
or, y = 6 - 2x Ans

Answer 3

For (-2,8), (1,1), and (3,k),
y=mx+c
A) 8 = m*(-2) + c
B) 1 = m*( 1) + c

Subtract A) from B)
-7 = m*3
m = -7/3

Plug back into B)
1 = (-7/3)*(1) + c
3/3 = -7/3 + c
c = 10/3.

y= (-7/3)*x + (10/3)
To find k in (3,k)
k = (-7/3)*3 + (10/3)
k= -21/3+10/3 = -11/3
------ ----- ----- ----- ------- -----
Yes, y = -2x + 6 is correct.

You can double check it real quick.
y = -2(9) + 6 = -12

x/3 + y/6 = 1
9/3 + (-12)/6 = 3 - 2 = 1.

It checks okay.

Answer 4

You're right about question 2.

For question 1, you'll need to know the equation of a line when you have two points on it.

When the two points are x1 y1 and x2 y2:

(y - y1)/(y2-y1) = (x - x1)/(x2 - x1)

This looks worse than it is. Just plug in the values you know for the first two points to get the equation of the line.

Now plug in the third point into the equation you've got and figure out k.

Answer 5

Question 1
m = (1 - 8) / 1 + 2 = - 7/3
m = (k - 1 ) / ( 3 -1) = (k - 1) / 2

(k - 1) / 2 = - 7/3 ------------- x both sides by 6
3(k - 1) = - 14
3k - 3 = - 14
3k = -11
k = -11/3

Question 2
x/3 + y/6 = 1--------------x both sides by 6
2x + y = 6
y = - 2x + 6 (the same as you!)