The equation of a straight line is `(2x)/(3)+(y)/(6)-5=0`
(i) Express the above equation in the form ax+b +c=0 and hence find the ordinate of a point whose abscissa is `(3)/(2)`
(ii) Express y in terms of x given that `(2x)/(3)+(y)/(6)-5=0` check whether (7,2) is a solution of given line .
(iii) Find the point where line `(2x)/(3)+(y)/(6)-5=0` cuts the y axis.

Explanation

(i) We have , `(2x)/(3)+(y)/(6)-5=0`
Multiplying both sides by 6 we get
4x+y -30 =0
`rArr` , 4x+(-30)=0
Which is of the form ax+by +c=-0 where a =4 =1 and c=-30 .
If abscissa x of a point (x,y) is `(3)/(2)` then to find the ordinate y , we put x =`(3)/(2)` in eq (1)
`(2)/(3)(3/2)+(y)/(6)=5 rArr (y)/(6)=5-1 rArr y=24`
When absicassa is `(3)/(2)` ordinate =24
(ii) `(2x)/(3)+(y)/(6)-5=0`
`(y)/(6)=5-(2x)/(3)`
`y=6(5-(x)/(3))`
y=30-4x
Put x =7 , y = 2 in eq . (2) we get
2-30-4(7) ltbvrgt 2=2 which is true
(7,2) lies on the given line
(iii) If a line cuts the y axis then x - coordinate is 0
so when x = 0, then on putting this value of x in the given equation
`(2x)/(3)+(y)/(6)-5=01`, we get y =30