Question

point of intersection of

3x/4 +y/2 +3/4=0; x/4 -3y/2 +2/3=0

Answer 1

Find the point of intersection of the two lines.
\[\frac{ 3x }{ 4 }+ \frac{ y }{2}+ \frac{ 3 }{ 4}=0 ; \frac{ x }{ 4 }- \frac{ 3y }{2 }+\frac{ 2 }{ 3 }=0\]

Answer 2

\[\frac{ 3x+2y+3 }{ 4}=\frac{ 3x-18y+8 }{12 }=0\]
\[12(3x+2y+3)=4(3x-18y+8)=0\]
\[36x+24y+36=12x-72y+32\]
left side:\[36x+24y+36\]\[>>>>>\] 1)\[y=\frac{- 3 }{ 2 }(x+1)\]
right side:\[12x-72y+32\]\[>>>>>>>>\]2)\[y=\frac{ 1 }{ 6 }x+\frac{ 4 }{9 }\]
so,\[\frac{ 1 }{ 6 }x+\frac{ 4 }{9 }=\frac{- 3 }{ 2 }(x+1)\] solving for x:
\[x=\frac{ -7 }{ 6 }\] substitute in 1)equation:\[y=\frac{ -3 }{ 2 }(\frac{ -7 }{6}+1)=\frac{ 1 }{ 4}\]
so, intersection point :\[(\frac{ -7 }{ 6 },\frac{ 1 }{ 4 })\]