Positioning a formula number

Fabrice Couvreur

31 Aug 2025 31 Aug '25

12:05 p.m.

Hi, In the example below, the formula number is not located on the baseline of the last equality. How can I make this happen ? Thanks for your help. Fabrice \starttext \startplaceformula \startformula \startalign \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR \NC \EQ a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR \stopalign \stopformula \stopplaceformula \stoptext

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Mikael Sundqvist

31 Aug 31 Aug

12:24 p.m.

Hi, On Sun, Aug 31, 2025 at 12:08 PM Fabrice Couvreur wrote:

...

Hi, In the example below, the formula number is not located on the baseline of the last equality. How can I make this happen ? Thanks for your help. Fabrice

\starttext \startplaceformula \startformula \startalign \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR \NC \EQ a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR \stopalign \stopformula \stopplaceformula \stoptext

Add it to the last \NR: \starttext \startplaceformula \startformula \startalign \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR \NC \EQ a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR[eq:foo] \stopalign \stopformula \stopplaceformula \stoptext /Mikael

Mikael Sundqvist

12:39 p.m.

Hi, or, if you prefer to "be modern", you can try \startformula f(x) \alignhere = a\left(x²+\frac{b}{a}x\right)+c \breakhere = a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c \breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c \breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a} \breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a} \breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a} \numberhere[eq:foo] \stopformula /Mikael On Sun, Aug 31, 2025 at 12:24 PM Mikael Sundqvist wrote:

...

Hi,

On Sun, Aug 31, 2025 at 12:08 PM Fabrice Couvreur wrote:

...
Hi, In the example below, the formula number is not located on the baseline of the last equality. How can I make this happen ? Thanks for your help. Fabrice

\starttext \startplaceformula \startformula \startalign \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR \NC \EQ a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR \stopalign \stopformula \stopplaceformula \stoptext

Add it to the last \NR:

\starttext \startplaceformula \startformula \startalign \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR \NC \EQ a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR[eq:foo] \stopalign \stopformula \stopplaceformula \stoptext

/Mikael

Fabrice Couvreur

1:40 p.m.

Hi Mikael, Thanks for your reply, and I'm not against a touch of modernity ! Fabrice Le dim. 31 août 2025 à 12:46, Mikael Sundqvist a écrit :

...

Hi,

or, if you prefer to "be modern", you can try

\startformula f(x) \alignhere = a\left(x²+\frac{b}{a}x\right)+c \breakhere = a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c \breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c \breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a} \breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a} \breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a} \numberhere[eq:foo] \stopformula

/Mikael

On Sun, Aug 31, 2025 at 12:24 PM Mikael Sundqvist wrote:

...
Hi,

On Sun, Aug 31, 2025 at 12:08 PM Fabrice Couvreur wrote:

...
Hi, In the example below, the formula number is not located on the

baseline of the last equality. How can I make this happen ?

...
...
Thanks for your help. Fabrice

\starttext \startplaceformula \startformula \startalign \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR \NC \EQ a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR \stopalign \stopformula \stopplaceformula \stoptext

Add it to the last \NR:

\starttext \startplaceformula \startformula \startalign \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR \NC \EQ a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR[eq:foo] \stopalign \stopformula \stopplaceformula \stoptext

/Mikael

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Fabrice Couvreur
Mikael Sundqvist