Finding a line perpendicular to a line and passing through the intersection of other two lines

Question

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

Answer 1

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:

$$y-\frac{22}{17}=-\frac{7}{6}(x+\frac{1}{17})\\[.1in]102y-128=-119x-1\\[.1in]102y+119x-127=0$$

Answer 2

The required answer is one of the members of the family [for some k to be determined later]

$$(2x − 3y + 4) + k(3x + 4y −5) = 0$$

Re-write it as $L_1 : (2 + 3k)x + (–3 + 4k)y + (4 – 5k) = 0$

such that $m(L_1) = \frac {– (2 + 3k)} {(4k – 3)}$

Given that $L_2 : 6x − 7y + c = 0$ with $m(L_2) = \frac {6}{7}$

‘$L_1$ is perpendicular to $L_2’$ implies $\frac {– (2 + 3k)}{(4k – 3)} \cdot \frac {6}{7} = –1$

∴ $k = 33/10$

Putting this value back in $L_1$ will give you the required result (which is $119x + 102y - 125 = 0$, machine confirmed)