How do you find equation of a line L which passes through the point $\left( {3, - 1} \right)$ and parallel to the line which passes through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$ ?
Answer
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Hint: Since the line is parallel to a given line passing through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$. So, the slope of the line is equal to the slope of the line passing through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$. Firstly, find the slope of line and put it in the general form of line that is $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$ where, $m$ is the slope of line and $\left( {{x_1},{y_1}} \right)$ is the point through which the line passes, to get the required equation of line.
Complete step by step answer:
Given: the line L passes through the point $\left( {3, - 1} \right)$ and is parallel to the line passing through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$.
We know that the slope of line passing through the points \[\left( {{x_1},{y_1}} \right)\] and $\left( {{x_2},{y_2}} \right)$ is given by $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
The slope of line passing through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$ is $\dfrac{{ - 3 - 5}}{{ - 2 - 0}} = \dfrac{{ - 8}}{{ - 2}} = 4$.
Since the line L is parallel to the line passing through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$. So, the slope of line L is $4$.
It is given that the line L is passing through $\left( {3, - 1} \right)$ and its slope is $4$. Putting these values into the general equation of line we get,
$
\Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) \\
\Rightarrow \left( {y - \left( { - 1} \right)} \right) = 4\left( {x - 3} \right) \\
\Rightarrow y + 1 = 4x - 12 \\
\Rightarrow 4x - y - 12 - 1 = 0 \\
\therefore 4x - y - 13 = 0 \\
$
Thus, the required equation of line L is $4x - y - 13 = 0$.
Note: The different form of the general equation of line are
(1) Two points form: If a line passing through two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$, then the equation of this line is given by $\left( {y - {y_1}} \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$.
(2) intercept form: If the slope of a line and its intercept on $y$-axis is given, then the equation of line is $y = mx + c$.
(3) Normal form: $X\cos \alpha + Y\sin \alpha = P$
Complete step by step answer:
Given: the line L passes through the point $\left( {3, - 1} \right)$ and is parallel to the line passing through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$.
We know that the slope of line passing through the points \[\left( {{x_1},{y_1}} \right)\] and $\left( {{x_2},{y_2}} \right)$ is given by $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
The slope of line passing through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$ is $\dfrac{{ - 3 - 5}}{{ - 2 - 0}} = \dfrac{{ - 8}}{{ - 2}} = 4$.
Since the line L is parallel to the line passing through the points $\left( {0,5} \right)$ and $\left( { - 2, - 3} \right)$. So, the slope of line L is $4$.
It is given that the line L is passing through $\left( {3, - 1} \right)$ and its slope is $4$. Putting these values into the general equation of line we get,
$
\Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) \\
\Rightarrow \left( {y - \left( { - 1} \right)} \right) = 4\left( {x - 3} \right) \\
\Rightarrow y + 1 = 4x - 12 \\
\Rightarrow 4x - y - 12 - 1 = 0 \\
\therefore 4x - y - 13 = 0 \\
$
Thus, the required equation of line L is $4x - y - 13 = 0$.
Note: The different form of the general equation of line are
(1) Two points form: If a line passing through two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$, then the equation of this line is given by $\left( {y - {y_1}} \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$.
(2) intercept form: If the slope of a line and its intercept on $y$-axis is given, then the equation of line is $y = mx + c$.
(3) Normal form: $X\cos \alpha + Y\sin \alpha = P$
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