Express $y$ in terms of $x$ , it begin given that $3 x - 2 y = 6$ check whether $(1, 2)$ is on the line represented by the equation $3 x - 2 y = 6$ . Draw its graph and also, find the value of $y$ at $x = 4$ .

y = 3

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y = 6

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y = 1

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y = 4

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Solution

The correct option is A $y = 3$
Given, equation: $3x - 2y = 6$.
Now, to check whether $(1, 2)$ will lie on the line $3 x - 2 y = 6$ , we have to put the value of $(x, y)$ in the given equation.
LHS $= 3 (1) - 2 (2) = 3 - 4 = - 1$ $\neq 6$ or RHS
Therefore, the point $(1, 2)$ will not lie on the line $3 x - 2 y = 6$ .
Now, at $x = 4$ , we have,
$3 (4) - 2 y = 6$
$y = 3$
Again, $3 x - 2 y = 6$
$\Rightarrow$ At $x = 0, y = - 3$ and at $y = 0, x = 2$
Therefore, table of points is:
$x$ $0$ $2$
$y$ $- 3$ $0$
So, graph of the given line is as:

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Q. Find the equation of the circle which circumscribes the triangle formed by the lines
(i) x + y + 3 = 0, x − y + 1 = 0 and x = 3
(ii) 2x + y − 3 = 0, x + y − 1 = 0 and 3x + 2y − 5 = 0
(iii) x + y = 2, 3x − 4y = 6 and x − y = 0.
(iv) y = x + 2, 3y = 4x and 2y = 3x.

Find the equation of the circle which circumscribes the triangle formed by the lines:
(i) $x + y + 3 = 0, x - y + 1 = 0$ and $x = 3$
(ii) $2 x + y - 3 = 0, x + y - 1 = 0$ and $3 x + 2 y - 5 = 0$
(iii) $x + y = 2, 3 x - 4 y = 6$ and $x - y = 0.$
(iv) $y = x + 2, 3 y = 4 x$ and $2 y = 3 x$ .