A student Ajay, was given a task to solve the pair of the equations x – y + 1 = 0 and
3x + 2y – 12 = 0 graphically. To draw the graph, he finds three solutions of each of the
given equations. After potting the points on graph paper, he draws the lines AB and PQ
representing the equation as shown in the following graph.
Based on the above graph, answer the following questions:
i. The equation of AB and PQ from the graph are:
a) AB : 3x + 2y – 12 = 0 & PQ : x – y + 1 = 0
b) AB : 3x + 2y – 12 = 0 & PQ : x – y – 1 = 0
c) AB : x – y + 1= 0 & PQ : 3x + 2y – 12 = 0
d) AB : x – y + 1= 0 & PQ : 3x – 2y – 12 = 0
ii. The solution of the pair of equation from the graph is :
a) x = 3, y = 2 b) x = 2 , y = 3 c ) x = 2, y = 2 d) x = 2, y = 4
iii. The coordinates of the vertices of the triangle formed by these lines and X-axis
are
a) (1,0) , (3,2) and (5,0) b) ( –2, 0), ( – 2,3) and (0,4)
c) (– 1, 0), ( 2, 3) and ( 4, 0) d) none of the above
iv. The area of the triangle formed by these lines and the Y-axis is :
a) 10 sq. units b) 5 sq. units c) 5
2
sq. units d) 2
5
sq. units
Answers
Given : Graph of pair of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0
To Find : The equation of AB and PQ from the graph are:
The solution of the pair of equation from the graph is :
The coordinates of the vertices of the triangle formed by these lines and X-axis
The area of the triangle formed by these lines and the Y-axis is :
Solution:
The equation of AB and PQ from the graph are:
AB = x – y + 1 = 0
PQ = 3x + 2y – 12 = 0
Hence correct option is c)
AB : x – y + 1= 0 & PQ : 3x + 2y – 12 = 0
The solution of the pair of equation from the graph is :
x = 2
y = 3
The coordinates of the vertices of the triangle formed by these lines and X-axis
(-1, 0) , ( 4 , 0) and ( 2 , 3)
area of the triangle formed by these lines and the Y-axis is :
= (1/2)(6 - 1) * 2
= 5 sq units
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Step-by-step explanation:
Ans= 5 sq units..
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