So here is the question as in my text book
Find the equation of the line through the intersection of $2x – 3y + 4 = 0 $ and $3x + 4y – 5 = 0$ and perpendicular to $6x – 7y + c = 0$
so I calculated the point of intersection of the two line $ 2x – 3y + 4 =0 $ and $3x + 4y – 5 = 0$ which turned out to be $ (-\frac{1}{17} , \frac{22}{17})$
Now since the line to be found is perpendicular to the line $ 6x – 7y + c =0 $ so used this formula to find the equation of the line
since they are perpendicular so product of their slopes will me $-1$ so i used this
.
let the line pass through point's $(x,y)$ and $(x_1, y_1) = (-\frac{1}{17} , \frac{22}{17})$ so i't slope will be $ \frac{y – y_1}{x-x_1} = m_1$ now slope of the other line let it be $m_2$ so $m_1*m_2 = -1$. So on solving this i got the equation of the line to be $139x + 42y – 125 = 0$
But this not the answer given in my textbook. Am i doing it in wrong way if i am then please a solution will help. Thank's
Best Answer
The slope of the line $6x-7y+c=0$ is $\frac{6}{7}$. This means that the slope of our desired line is $-\frac{7}{6}$.
Using the point-slope form of a line, we have:
$$y-\frac{22}{17}=-\frac{7}{6}(x+\frac{1}{17})\\[.1in]102y-128=-119x-1\\[.1in]102y+119x-127=0$$