SOLUTION: a straight line with equation 2y = kx + h passes through the points (-3,6) and ( 1,11) find the value of k and of h

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Question 1160656: a straight line with equation 2y = kx + h passes through the points (-3,6) and ( 1,11) find the value of k and of h

Found 4 solutions by josgarithmetic, MathLover1, MathTherapy, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
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system%282%2A6=-3k%2Bh%2C2%2A11=1%2Ak%2B11%29

Simplify and solve the system for k and h.

Answer by MathLover1(20850) About Me  (Show Source):
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2y+=+kx+%2B+h passes through the points (-3,6) and ( 1,11)
find the value of k and of+h

2y+=+kx+%2B+h use the point (-3,6)
2%2A6+=+k%28-3%29+%2B+h........solve for h
12+=+-3k+%2B+h
h=3k%2B12............eq.1
use the point ( 1,11)
2%2A11+=+k%281%29+%2B+h.....solve for h
22+=+k+%2B+h
h=22-k............eq.2

from eq.1 and eq.2 we have:
3k%2B12=22-k
3k%2Bk=22-12
4k=10
k=10%2F4
k=5%2F2
go to h=22-k............eq.2, plug in k
h=22-5%2F2
h=39%2F2
then your equation is:

2y+=+%285%2F2%29x+%2B+39%2F2
and in slope intercept form is:
y+=+%285%2F4%29x+%2B+39%2F4




Answer by MathTherapy(10552) About Me  (Show Source):
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a straight line with equation 2y = kx + h passes through the points (-3,6) and ( 1,11) find the value of k and of h
2y = kx + h
2(6) = - 3k + h ---- Substituting (- 3, 6) for (x, y)
12 = - 3k + h ------ eq (i)

2y = kx + h
2(11) = k + h --- Substituting (1, 11) for (x, y)
22 = k + h ------ eq (ii)
10 = 4k --------- Subtracting eq (i) from eq (ii)
highlight_green%28matrix%281%2C5%2C+10%2F4%2C+or%2C+5%2F2%2C+%22=%22%2C+k%29%29 

matrix%281%2C3%2C+22%2C+%22=%22%2C+5%2F2+%2B+h%29 ----- Substituting 5%2F2 for k in eq (ii)
matrix%282%2C3%2C+22+-+5%2F2%2C+%22=%22%2C+h%2C+44%2F2+-+5%2F2%2C+%22=%22%2C+h%29
highlight_green%28matrix%281%2C5%2C+39%2F2%2C+or%2C+19%261%2F2%2C+%22=%22%2C+h%29%29

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

1)  k%2F2  is the slope, and it can be easily calculated via the coordinates of the given points


        k%2F2 = %2811-6%29%2F%281-%28-3%29%29 = 5%2F4;  so   k = 5%2F2.


    You DO NOT NEED to solve system of equation to find k.


2)  As soon as you know "k", substitute it and the coordinates of any of the two point into the given equation to find "h"


        2*11 = %285%2F2%29%2A1 + h,

        h = 22 - 2.5 = 19.5.

Solved.

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To solve the problem in full, there is NO NEED to solve any system of equations.

It is the major lesson to learn from this assignment.