reduce the line 2x-3y+5=0,in slope intercept intercept and normal forme

slope intercept form is y=mx+c .. the equation of line given by u can b converted into the same form by just rearrainging the equation !!

normal form is xcosQ+ysinQ=p (here Q stands for angle made by the line ) which can be calculated by the above mentioned equation by finding the value of m !! from y=mx+c , m=tanQ . whereas p stands for the perpendicular distance to the line y=mx+c from the origion !!

2x-3y+5=0

→y=(2/3)x+(5/3)

here slope=2/3 and y-intercept is 5/3

This is slope-intercept form.

2x-3y+5=0

→2x-3y=-5

→(-2/5)x+(3/5)y=1

→x/(-2/5)+y/(5/3)=1

here,x-intercept=(-2/5) and y-intercept=5/3

This is intercept form.

2x-3y+5=0

=>(-2/5)x+(3/5)y=1

=>x/(-5/2)+y/(5/3)=1.........................intercept form

2x-3y+5=0

=>(2/13^0.5)x-(3/13^0.5)y+13^-0.5=0

=>(-2/13^0.5)x+(3/13^0.5)y=13^-0.5..............normal form