robertor7791
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find an equation for the line

Find An Equation For The Line class=

Sagot :

Answer:  [tex]y= \frac{3}{5} x+\frac{25}{18}[/tex]   or [tex]y+1=\frac{3}{5}\left(x+4)[/tex]  or [tex]y-5=\frac{3}{5}\left(x-6)[/tex]

All three equations represent the same function.

Step-by-step explanation:

Point-slope form of a line:  [tex]y-y_{1}=m\left(x-x_{1}\right)[/tex]

Let     [tex](x_1,y_1)\\[/tex]   =   (-4, -1)

Let     (x, y)      =   (6, 5)

[tex]5-(-1)=m (6-(-4))[/tex]

Solve for m:

[tex]6=m (10)\\\frac{3}{5} =m[/tex]

Plug into slope intercept form y=mx+b

[tex]y=\frac{3}{5} x+b[/tex]

Re-use (x, y) to find b

[tex]5 = (3/5)*6+b\\b= 25/18[/tex]

If you wanted point-slope form, just re-use either of the points above to plug for [tex](x_1,y_1)\\[/tex].

Esther

Answer:

[tex]y=\frac{3}{5}x+\frac{7}{5}[/tex]

Step-by-step explanation:

Slope (m) formula: [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

[tex]m=\frac{y_2-y_1}{x_2-x_1} \\\\m=\frac{5-(-1)}{6-(-4)} \\\\m=\frac{5+1}{6+4} \\\\m=\frac{6}{10} \\\\m=\frac{3}{5}[/tex]

[tex]y=mx+b[/tex]

[tex]y=\frac{3}{5} x+b[/tex]

(-4, -1)

[tex]y=\frac{3}{5} x+b\\\\-1=\frac{3}{5}(-4)+b\\\\-1=-\frac{12}{5}+b\\\\5(-1=-\frac{12}{5}+b)\\\\-5=-12+5b\\\\+12 +12\\\\7=5b\\\\/5 /5\\\\7/5=b[/tex]

[tex]y=\frac{3}{5}x+\frac{7}{5}[/tex]

Hope this helps!

View image Esther
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