Question 1203163: The straight line y=mx+c passes through the points (0, k) and (h,2k)
Express m and c in terms of h and k
What I have tried
A. B.
.........................
(0,k) (h, 2k)
(X1,Y1). (X2, Y2)
M= (Y2-Y1)/(X2-X1)
M= k/h
Y-Y1=m(X-X1)
====
Y=kx/h +K
C=K
Found 3 solutions by Alan3354, josgarithmetic, MathLover1:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The straight line y=mx+c passes through the points (0, k) and (h,2k)
Express m and c in terms of h and k
What I have tried
A. B.
.........................
(0,k) (h, 2k)
(X1,Y1). (X2, Y2)
M= (Y2-Y1)/(X2-X1)
M= k/h
Y-Y1=m(X-X1)
====
Y=kx/h +K
C=K
===============================
Slope m = (2k-k)/(h-0) = k/h
---------------
(0,k) is the y-intercept, so c = k
Answer by josgarithmetic(39620)  (Show Source):
You can put this solution on YOUR website! The two points (0,k) and (h, 2k);
Directly to formula of slope,
Also using slope intercept form,  and now having an expression found for slope m,
picking your first given point (0,k),
 , which you could have jumpted through much more quickly;
Understand, first given point (0,k) indicates that vertical axis intercept is k.
You would not really need to SOLVE for c.
Answer by MathLover1(20850)  (Show Source):
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