How do you write an equation of the line passing through \[\left( { - 4, - 5} \right)\] and slope\[\dfrac{{ - 3}}{2}\]?
Answer
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Hint: In this question we have to asked to find the equation, where slope and the point through which the equation passing are given, we will make use of the formula \[y - {y_1} = m\left( {x - {x_1}} \right)\], where m is the slope and \[\left( {{x_1},{y_1}} \right)\] is the point through which the equation passes, now substituting the given values in the formula we will get the required equation.
Complete step-by-step answer:
Given a line passing through the point \[\left( { - 4, - 5} \right)\] and slope \[\dfrac{{ - 3}}{2}\], and we have the equation.
So we will use the formula \[y - {y_1} = m\left( {x - {x_1}} \right)\], where m is the slope and \[\left( {{x_1},{y_1}} \right)\] is the point through which the equation passes,
Now here,\[{x_1} = - 4\],\[{y_1} = - 5\]and\[m = \dfrac{{ - 3}}{2}\],
By substituting the values in the formula we get,
\[ \Rightarrow y - \left( { - 5} \right) = \dfrac{{ - 3}}{2}\left( {x - \left( { - 4} \right)} \right)\],
Now simplifying we get,
\[ \Rightarrow y + 5 = \dfrac{{ - 3}}{2}\left( {x + 4} \right)\],
Now multiplying 2 to the both sides of the equation we get,
\[ \Rightarrow \left( {y + 5} \right) \times 2 = \dfrac{{ - 3}}{2}\left( {x + 4} \right) \times 2\],
Now simplifying the equation we get,
\[ \Rightarrow 2\left( {y + 5} \right) = - 3\left( {x + 4} \right)\],
Now multiplying for opening the brackets we get,
\[ \Rightarrow 2y + 10 = - 3x - 12\],
Now taking all terms to one side we get,
\[ \Rightarrow 2y + 10 + 3x + 12 = 0\],
Now adding the like terms we get,
\[ \Rightarrow 3x + 2y + 22 = 0\].
So, the required equation is \[3x + 2y + 22 = 0\].
\[\therefore \]The equation which passes through the point \[\left( { - 4, - 5} \right)\] and slope \[\dfrac{{ - 3}}{2}\] is equal to \[3x + 2y + 22 = 0\].
Note:
Linear equations are straight lines equations that have simple variables expressions with terms without exponents on them. There are many methods to find the equation of a line in two variables. We will use the slope-point in the questions like the given one. And other methods are slope-intercept form where a slope and y-intercept are given i.e.,\[y = mx + c\], intercept form where x-intercept and y-intercept are given i.e.,\[\dfrac{x}{a} + \dfrac{y}{b} = 1\], and two point’s form where two points through which line passes through them will be given, i.e.,\[\dfrac{{y - {y_1}}}{{{y_2} - {y_2}}} = \dfrac{{x - {x_1}}}{{{x_2} - {x_1}}}\].
Complete step-by-step answer:
Given a line passing through the point \[\left( { - 4, - 5} \right)\] and slope \[\dfrac{{ - 3}}{2}\], and we have the equation.
So we will use the formula \[y - {y_1} = m\left( {x - {x_1}} \right)\], where m is the slope and \[\left( {{x_1},{y_1}} \right)\] is the point through which the equation passes,
Now here,\[{x_1} = - 4\],\[{y_1} = - 5\]and\[m = \dfrac{{ - 3}}{2}\],
By substituting the values in the formula we get,
\[ \Rightarrow y - \left( { - 5} \right) = \dfrac{{ - 3}}{2}\left( {x - \left( { - 4} \right)} \right)\],
Now simplifying we get,
\[ \Rightarrow y + 5 = \dfrac{{ - 3}}{2}\left( {x + 4} \right)\],
Now multiplying 2 to the both sides of the equation we get,
\[ \Rightarrow \left( {y + 5} \right) \times 2 = \dfrac{{ - 3}}{2}\left( {x + 4} \right) \times 2\],
Now simplifying the equation we get,
\[ \Rightarrow 2\left( {y + 5} \right) = - 3\left( {x + 4} \right)\],
Now multiplying for opening the brackets we get,
\[ \Rightarrow 2y + 10 = - 3x - 12\],
Now taking all terms to one side we get,
\[ \Rightarrow 2y + 10 + 3x + 12 = 0\],
Now adding the like terms we get,
\[ \Rightarrow 3x + 2y + 22 = 0\].
So, the required equation is \[3x + 2y + 22 = 0\].
\[\therefore \]The equation which passes through the point \[\left( { - 4, - 5} \right)\] and slope \[\dfrac{{ - 3}}{2}\] is equal to \[3x + 2y + 22 = 0\].
Note:
Linear equations are straight lines equations that have simple variables expressions with terms without exponents on them. There are many methods to find the equation of a line in two variables. We will use the slope-point in the questions like the given one. And other methods are slope-intercept form where a slope and y-intercept are given i.e.,\[y = mx + c\], intercept form where x-intercept and y-intercept are given i.e.,\[\dfrac{x}{a} + \dfrac{y}{b} = 1\], and two point’s form where two points through which line passes through them will be given, i.e.,\[\dfrac{{y - {y_1}}}{{{y_2} - {y_2}}} = \dfrac{{x - {x_1}}}{{{x_2} - {x_1}}}\].
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