Question

How do you write an equation in standard form with integer coefficients for the line with slope \[\dfrac{16}{11}\] going through the point (-6, -1)?

Answer 1

Hint: First find the equation of the line in slope – intercept form given as: - \[y=mx+c\], where ‘m’ is the slope and ‘c’ is the intercept on the y – axis. Substitute the value of the given point (-6, -1) and find the linear equation obtained and convert it in the form \[ax+by+c'=0\] to get the answer.

Complete step by step answer:
Here, we have been provided with the information that a line has slope \[\dfrac{16}{11}\] and is passing through the point (-6, -1). We have been asked to write the equation in standard form.
Now, first let us find the equation of the line in slope – intercept form then we will convert it into the standard form. We know that the equation of line in slope – intercept form is given as: - \[y=mx+c\], where ‘m’ is the slope and ‘c’ is the intercept on the y – axis. Since, the slope of this line is given as \[\dfrac{16}{11}\], so we have,
\[\Rightarrow y=\dfrac{16}{11}x+c\]
Now, it is given that this line is passing through the point (-6, -1) that means this line will satisfy the point (-6, -1). So, substituting the values of x and y, we get,
\[\begin{align}
  & \Rightarrow -1=\dfrac{16}{11}\times \left( -6 \right)+c \\
 & \Rightarrow -1=\dfrac{96}{11}+c \\
 & \Rightarrow c=\dfrac{96}{11}-1 \\
 & \Rightarrow c=\dfrac{96-11}{11} \\
 & \Rightarrow c=\dfrac{85}{11} \\
\end{align}\]
Therefore, the equation of the line in slope – intercept form is given as: -
\[\Rightarrow y=\dfrac{16x}{11}+\dfrac{85}{11}\] - (1)
We know that the equation of a line in standard form is given as \[ax+by+c'=0\], where ‘a’ is the coefficient of x, ‘b’ is the coefficient of ‘y’ and c’ is the constant term. Now, multiplying both sides of the equation (1) with 11, we get,
\[\Rightarrow 11y=16x+85\]
Taking all the terms to the L.H.S., we get,
\[\Rightarrow 11y-16x-85=0\]
On comparing the above equation with \[ax+by+c'=0\], we have,
\[\Rightarrow \] a = -16
\[\Rightarrow \] b = 11
\[\Rightarrow \] c’ = -85
Clearly, we can see that a, b and c’ are all integers. Hence, the required equation of the line is \[-16x+11y-85=0\].

Note:
One may note that here we can directly derive the standard form without deriving the intercept form of the given line but in that case we would have to find the values of a, b and c’ separately by solving three linear equations. This is the reason that we have derived the slope – intercept form first. You must remember all the forms and their general equations for a straight line, like: - intercept form, slope – intercept form, point – slope form, standard form, polar form etc. Remember that these forms are easily interconvertible.