2.) Use the Slope Intercept Form of a line to find the equation of the line from point C to point D.


Slope Intercept Form of a Line:

y = mx + b

m is the slope and b is the y-intercept

2 Use the Slope Intercept Form of a line to find the equation of the line from point C to point D Slope Intercept Form of a Line y mx b m is the slope and b is class=

Respuesta :

First we need slope

  • C=(0,0)
  • D(7,12)

[tex]\\ \sf\longmapsto m=\dfrac{12-0}{7-0}[/tex]

[tex]\\ \sf\longmapsto m=\dfrac{12}{7}[/tex]

Put D co-ordinates on y=mx+b

[tex]\\ \sf\longmapsto 12=\dfrac{12}{7}(7)+b[/tex]

[tex]\\ \sf\longmapsto 12=12+b[/tex]

[tex]\\ \sf\longmapsto b=12-12[/tex]

[tex]\\ \sf\longmapsto b=0[/tex]

Now

slope intercept form.

[tex]\\ \sf\longmapsto y=\dfrac{12}{7}x[/tex]

  • As b=0

Answer:

• General equation of a line:

[tex]{ \rm{y = mx + b}}[/tex]

Consider end points of line CD: (0, 0) and (7, 12):

• find the slope, m:

[tex]{ \rm{slope = \frac{y _{2} - y _{1} }{x _{2} - x _{1} } }} \\ \\ { \tt{m = \frac{12 - 0}{7 - 0} }} \\ \\ { \underline{ \tt{ \: \: m = \frac{12}{7} \: \: }}}[/tex]

• using end point (0, 0), find the y-intercept, b:

[tex]{ \rm{y = mx + b}} \\ \\ { \rm{0 = ( \frac{12}{7} \times 0) + b}} \\ \\ { \tt{b = 0}}[/tex]

• therefore, equation of line CD is;

[tex]{ \boxed{ \boxed{ \tt{ \: \: y = \frac{12}{7}x \: \: }}}}[/tex]

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