Original question
The solution of the equation $$ \frac{dy}{dx}= \frac{3x-4y-1}{3x-4y-3}$$ passes through the origin. The distance of it from the point (-1,1) is ___.
*My attempt *
Found the solution of DE as $$ x-y=ln(3x-4y+1)$$. But how I will find the distance from the given point?
Edit-
Found the normal as $$ (h+1)y=(k-1)x+(h+k)$$ where (h, k) lies on the curve for the intersection point. But how to find the intersection point value? Its very tough to handle the equation. Any hint will be helpful.
Hint: You have to find the equation of the normal to the curve passing through the point $(-1,1)$
After finding the equation of the normal you can easily find the intersection point between the curve and the normal, then apply the distance formula between two points.
Edit:
To find normal you now that $$m_1\cdot m_2=-1$$
Thus the slope of normal is $$\frac{-1}{\frac{3x-4y-1}{3x-4y-3}}=-\frac{3x-4y-3}{3x-4y-1}$$
The equation of normal at some point $P(x_0,y_0)$ is
$$y-y_0=-\frac{3x_0-4y_0-3}{3x_0-4y_0-1}(x-x_0)$$
Also $P(x_0,y_0)$ satisfies the curve thus $$x_0-y_0=ln(3x_0-4y_0+1)$$
Also the normal passes through the point$(-1,1)$ thus
$$1-y_0=-\frac{3x_0-4y_0-3}{3x_0-4y_0-1}(-1-x_0)$$
can you solve for $x_0 \text{and} y_0$