Question

How do you write an equation in standard form with integer coefficients for the line with slope $\dfrac{{17}}{{12}}$ going through the point (-5,-3)?

Answer 1

Hint : Taking out the initial data from the question, we will use the general equation of a straight line which is used when we know one point and slope of it $(y - {y_1}) = m(x - {x_1})$ . Where m is the slope, ${x_1}\,$ and ${y_1}$ are the points on it. Putting the values from the question, we will bring like terms on one side and unknowns on the other. After that, we will simplify the equation which will in the end give us an answer of the general form $ax + by = c$ .
Use the formula $(y - {y_1}) = m(x - {x_1})$

Complete step-by-step answer :
The given data from the question is
$m = \dfrac{{17}}{{12}}$ and ${x_1} = - 5$ and ${y_1} = - 3....................(1)$
We know the general formula of a straight line when a slope and a point on it is given by:-
$(y - {y_1}) = m(x - {x_1})....................(2)$
Putting the values from (1) to (2) we will get
$ax + by = c$
Solving the equation further we will get

$$(y + 3) = \dfrac{{17}}{{12}}(x + 5) \\\ \Rightarrow y + 3 = \dfrac{{17}}{{12}}x + \left( {\dfrac{{17 \times 5}}{{12}}} \right) \\\ \Rightarrow y + 3 = \dfrac{{17}}{{12}}x + \dfrac{{85}}{{12}} \;$$

Bringing all the constants on one side and the unknown variables on one side of the equation, we will get:-
$y + 3 = \dfrac{{17}}{{12}}x + \dfrac{{85}}{{12}} \\\ \Rightarrow y - \dfrac{{17}}{{12}}x = \dfrac{{85}}{{12}} - 3 \\\ \Rightarrow y - \dfrac{{17}}{{12}}x = \dfrac{{85 - 36}}{{12}} \\\ \Rightarrow y - \dfrac{{17}}{{12}}x = \dfrac{{49}}{{12}} \;$
Multiplying the whole equation by 12 we will get
$\left( {y \times 12} \right) - \left( {\dfrac{{17}}{{12}} \times x \times 12} \right) = \dfrac{{49}}{{12}} \times 12 \\\ \Rightarrow 12y - 17x = 49 \\\ \Rightarrow 17x - 12y = - 49 \;$
Therefore, a line with slope $\dfrac{{17}}{{12}}$ and passing through (-5,-3) will have the equation $17x - 12y = - 49$ .
So, the correct answer is “ $17x - 12y = - 49$ ”.

Note : While putting the values in equation (2) one must take care of the sign of the initial equation and the point both. Like here, the equation itself had negative signs in it, and the point was also negative. Therefore, negative negative became positive. So, make sure you use the correct sign as this is a very simple and easy to avoid mistake.