Answer :

The given equation is 5x – 3y + 30 = 0

(i) Given A ( – 6,0). Here x = – 6 and y = 0


On substituting x = – 6, y = 0 in LHS of given equation, we get


LHS = 5( – 6) – 3(0) + 30 = – 30 + 30 = 0 = RHS


So, x = – 6, y = 0 is a solution of the equation 5x – 3y + 30 = 0.


Hence, point A lies on the graph of the linear equation 5x – 3y + 30 = 0.


(ii) Given B (0,10). Here x = 0 and y = 10


On substituting x = 0, y = 10 in LHS of given equation, we get


LHS = 5(0) – 3(10) + 30 = – 30 + 30 = 0 = RHS


So, x = 0, y = 10 is a solution of the equation 5x – 3y + 30 = 0


Hence, point B lies on the graph of the linear equation 5x – 3y + 30 = 0.


(iii) Given C (3, – 5). Here x = 3 and y = – 5


On substituting x = 3, y = – 5 in LHS of given equation, we get


LHS = 5(3) – 3( – 5) + 30 = 15 + 15 + 30 = 60 ≠ RHS


So, x = 3, y = – 5 is not a solution of the equation 5x – 3y + 30 = 0


Hence, point C does not lie on the graph of the linear equation 5x – 3y + 30 = 0.


(iv) Given D (4,2). Here x = 4 and y = 2


On substituting x = 4, y = 2 in LHS of given equation, we get


LHS = 5(4) – 3(2) + 30 = 20 – 6 + 30 = 44 ≠ RHS


So, x = 4, y = 2 is not a solution of the equation 5x – 3y + 30 = 0


Hence, point D does not lie on the graph of the linear equation 5x – 3y + 30 = 0.


(v) Given E ( – 9,5). Here x = – 9 and y = 5


On substituting x = – 9, y = 5 in LHS of given equation, we get


LHS = 5( – 9) – 3(5) + 30 = – 45 – 15 + 30 = – 30 ≠ RHS


So, x = – 9, y = 5 is not a solution of the equation 5x – 3y + 30 = 0


Hence, point E does not lie on the graph of the linear equation 5x – 3y + 30 = 0.


(vi) Given F ( – 3,5). Here x = – 3 and y = 5


On substituting x = – 3, y = 5 in LHS of given equation, we get


LHS = 5( – 3) – 3(5) + 30 = – 15 + 15 + 30 = 0 = RHS


So, x = – 3, y = 5 is a solution of the equation 5x – 3y + 30 = 0


Hence, point F lies on the graph of the linear equation 5x – 3y + 30 = 0.


(vii) Given G ( – 9, – 5). Here x = 3 and y = – 5


On substituting x = – 9, y = – 5 in LHS of given equation, we get


LHS = 5( – 9) – 3( – 5) + 30 = – 45 + 15 + 30 = 0 = RHS


So, x = – 9, y = – 5 is a solution of the equation 5x – 3y + 30 = 0


Hence, point G lies on the graph of the linear equation 5x – 3y + 30 = 0.


Or Graphically


Here, we can see through the graph also that Point A, B, F and G lie on the graph of the linear equation 5x – 3y + 30 = 0



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