persamaan garis yang melalui titik A (1,4) tegak lurus dengan 2X - 5Y = 0

g 2x-5y =
(-2/-5) = 2/5

tegak lurus m1 xm2= -1
2/5 xm2= -1
m2= -5/2


persamaan y-y1=m(x-x1)
y-4=-5/2(x-1)
2y-8=-5(x-1)
2y-8=-5x+5
5x+2y=13

[tex]2x - 5y = 0 \\ - 5y = - 2x \\ y = \frac{2}{5} x \\ maka \: m = \frac{2}{5} \\ krna \: tegak \: lurus \: m1.m2 = - 1 \\ \frac{2}{5} \times m2 \ = - 1 \\ m2 = - 1 \times \frac{5}{2} = - \frac{5}{2} \\ \\ melalui \: (1.4) \\ y - y1 = m( x - x1) \\ y - 4 = - \frac{5}{2}(x - 1) \\ 2y - 8 = - 5x + 5 \\ 2y = - 5x + 13 \\ y = - \frac{ 5}{2} x + \frac{13}{2} [/tex]