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y-y1=m(x-x1)
y-(-3)=3/5(x-2)
y+3=3/5(x-2)
y=3/5x-6/5-3
5y=3x-6-15
5y=3x-21
atau 3x-5y = 21
3y = -5x+12
y = => Jadi, gradien garisnya = m1 =
Jadi gradien garis titik (2,-3) adalah = m2 = (didapat dari : gradien tegak lurus = m1 x m2 =-1)
Persaman garis titik (2,-3) =
y-y1 = m (x-x1)
y - (-3) = - (x-2)
y+3 =
5(y+3) = 5 ( )
5y+15 = 3x -6
-3x + 5y +15 +6 = 0
-3x + 5y +21 =0
Jadi persamaan garis melalui titik (2,-3) adalah-3x+5y+21 = 0 atau 3x-5y-21 = 0