How do you graph a line written in Standard Form?

  find the intercepts and draw the line through them

Step-by-step explanation:

A standard form linear equation looks like ...

  ax +by = c

where a, b, c are mutually prime and the leading coefficient is positive. (For a ≠ 0, a is the leading coefficient; for a=0, b is the leading coefficient.)

Except in the special cases of ...

the line will intercept both axes somewhere. Those intercepts are easily found.

Setting x = 0 finds the y-intercept ...

  a·0 +by = c

  y = c/b . . . . . . . y-intercept

Setting y = 0 finds the x-intercept ...

  ax +b·0 = c

  x = c/a . . . . . . . x-intercept

Plot the two intercepts and draw the line through them.

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If one of the intercepts is an integer, and the other is not, it may be of interest to plot the integer intercept and use the slope to find another point. The slope is ...

  m = -a/b . . . . . . slope of a standard-form line

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Additional comment

With a little practice, it can become easy to read the intercepts from the equation. This knowledge can help you visualize the quadrants the line passes through, and can give you some idea of its slope. As a consequence of that visualization, you can get an idea of where the solution to a system of equations lies: which quadrant, and where in that quadrant, relative to the intercept values.

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Example

  3x +4y = 12

The y-intercept (x=0) is 12/4 = 3.

The x-intercept (y=0) is 12/3 = 4.

These are both on the positive axes, so the slope of the line is negative. The triangle formed with the axes is in the first quadrant. (see attached)

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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step-by-step explanation: