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प्रश्न
Find graphically the vertices of the triangle, whose sides are given by 3y = x + 18, x + 7y = 22 and y + 3x = 26.
उत्तर
The given equation are :
3y = x + 18 ...(1)
x + 7y = 22 ...(2)
y + 3x = 26 ...(3)
3y = x + 18
⇒ x = 3y - 18
Corresponding values of x and y can be tabulated as :
x | 0 | -3 | -6 |
y | 6 | 5 | 4 |
Plotting points (0, 6), (-3, 5) and (-6, 4) joining them, we get a line l1 which is the graph of equation (1).
Again, x + 7y = 22
⇒ x = 22 - 7y
Corresponding values of x and y can be tabulated as :
x | 1 | 8 | -6 |
y | 3 | 2 | 4 |
Plotting points (1, 3), (8, 2), (-6, 4) and joining them, we get a line l2 which is the graph of equation (2).
Also, y + 3x = 26
⇒ y = 26 - 3x
Corresponding values of x and y can be tabulated as :
x | 7 | 8 | 9 |
y | 5 | 2 | -1 |
Plotting points (7, 5), (8, 2), (9, -1) and joining them, we get a line l3 which is the graph of equation (3).
It can be seen that the lines l1, l2 and l3 intersect each other to form triangle ABC.
The verticles of ΔABC are A(-6, 4), B(6, 8) and C(8, 2).
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