Answered

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(-3,1) and (2,-2)
linear equations
how to solve it?


Sagot :

Finding first the slope of the line you'll have:
[tex]m = \frac {y_2-y_1}{x_2-x_1}[/tex]
[tex]m = \frac{-2-1}{2-(-3)} [/tex]
[tex]m = -\frac{3}{5}[/tex]
-------------------------------
Using intercept form
y - y1 = m(x-x1)
taking point (2,-2)
y - (-2) = -3/5 (x-2)
y + 2 = -3/5 (x-2)
cross multiply 5
5(y+2) = -3(x-2)
5y + 10 = -3x + 6
5y + 3x = 6 - 10
5y + 3x = -4    ----in standard form

5y = -3x - 4
y = -3/5x - 4/5   ---in y-intercept form
First find the slope
(-2-1)(2+3)
=-3/5

Use the formula y-y1=m(x-x1)
5[y-1=-3/5(x+3)]
5y-5=-3x-9
3x+5y=-4

The equation is 3x+5y=-4