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Question: Find the equation of the line parallel to \({\text{3x + 2y = 8}}\) and passing through (0, 1)....

Find the equation of the line parallel to 3x + 2y = 8{\text{3x + 2y = 8}} and passing through (0, 1).

Explanation

Solution

We first find the slope (m) of the given equation. Note that the slope of a line parallel to the given line will also have the same value.
So, we convert the equation of the given line in the form of y=mx+cy = mx + c, where m is the slope.
So, the line parallel to the given line will be of the form y=mx+c1y = mx + {c_1}, where we need to find c1{c_1}.
Now, given that the required line passes through (0, 1)
Therefore, find the value of c1{c_1} by substituting x=0 and y=1 in its equation. Thus we get the required equation of line.

Complete step-by-step answer:
Consider the given equation 3x + 2y = 8{\text{3x + 2y = 8}}
2y=83x\Rightarrow 2y = 8 - 3x
On dividing the equation by 2 we get,
y=32x+4\Rightarrow y = - \frac{3}{2}x + 4
So, this equation is in the form of y=mx+cy = mx + c, where m is the slope.
∴ Slope of the given line is 32 - \frac{3}{2}
∴ The slope of the required line is also 32 - \frac{3}{2} , since it is parallel to the given line, and as parallel lines have the same slope.
Let, the equation of this line be y=32x+c1(1)y = - \frac{3}{2}x + {c_1} \ldots (1)
Given that, the line passes through (0, 1)
∴ Substituting x=0 and y=1 in equation (1), we get,
1=(32×0)+c11 = \left( { - \frac{3}{2} \times 0} \right) + {c_1}
On simplification we get,
c1=1\Rightarrow {c_1} = 1
Therefore,
y=32x+c1 y=32x+1 \begin{gathered} y = - \frac{3}{2}x + {c_1} \\\ \Rightarrow y = - \frac{3}{2}x + 1 \\\ \end{gathered}
On multiplying the equation by 2 we get,
2y=23x\Rightarrow 2y = 2 - 3x
On rearranging the terms we get,
3x+2y=2\Rightarrow 3x + 2y = 2
Hence, the equation of the line parallel to 3x + 2y = 8{\text{3x + 2y = 8}} and passing through (0, 1) is 3x+2y=23x + 2y = 2 .

Note: (Alternative method): The equation of a straight line with slope m and passing through the point (x1, y1) is given by
yy1=m(xx1)y - {y_1} = m(x - {x_1})
You can directly substitute the values of m, x1 and y1 here to find the equation of the required line.