2x−3y−5/4=02x-3y- 5/4 =02x−3y−5/4=0
y=8x−512y=\dfrac{8x-5}{12}y=128x−5
The y- intercept of the line in point A(0,y1)A(0,y_1)A(0,y1)
Consider A(0;y1)A(0;y_1)A(0;y1) in y=8x−512y=\dfrac{8x-5}{12}y=128x−5
y1=8∗0−512=−512y_1=\dfrac{8*0-5}{12}=-\dfrac{5}{12}y1=128∗0−5=−125
A(0,−512)A(0,-\dfrac{5}{12})A(0,−125)
The x- intercept of the line in point B(x1,0)B(x_1,0)B(x1,0)
Consider B(x1,0)B(x_1,0)B(x1,0) in y=8x−512y=\dfrac{8x-5}{12}y=128x−5
0=x1∗8−5120=\dfrac{x_1*8-5}{12}0=12x1∗8−5
x1∗8−5=0x_1*8-5=0x1∗8−5=0
x1=58x_1=\dfrac{5}{8}x1=85
B(58,0)B(\dfrac{5}{8},0)B(85,0)
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