Find the equation of line in new co-ordinate system.

Question

A line is represented by equation $4x+5y=6$ in the co-ordinate system with the origin $(0,0)$.You are required to find the equation of the straight line perpendicular to this line that passes through the point $(1,-2)$ [which is in the co-ordinate system in which the origin lies in $(-2,-2)$].

$\color{green}{a.)\ 5x-4y=11}\\ b.)\ 5x-4y=13\\ c.)\ 5x-4y=-3\\ d.)\ 5x-4y=7$

$\quad$

$4(x-2)+5(y-2)=6\\ \implies 4x+5y=24\\ \text{slope}_1=-\dfrac{4}{5}\\ y+2=\dfrac{5}{4}(x-1)\\ 5x-4y=13$

But the book is giving as option $a.)$

I look for short/simple method.

I have studied maths up to $12th$ grade.

Answer 1

$$ 5x-4y=13 $$ It would have been the equation of line if the origin would not have been shifted.

As the origin is now shifted to $(-2,-2)$ , so $x$ has now become $x+2$ and $y$ has become $y+2$. On substituting new values of $x$ and $y$, $$ 5(x+2)-4(y+2)=13 $$ $$ 5x-4y=13-2 $$ $$ 5x-4y=11 $$

Answer 2

The equation of line passing through the point $(1, -2)$ & normal to the line: $4x+5y=6$ (having slope $\frac{-4}{5}$) is given as follows (for the origin $(0, 0)$) $$y-(-2)=\frac{5}{4}(x-1) $$$$\implies 5x-4y-13=0$$ Now, shift the origin $(0, 0)$ to the given point $(-2, -2)$ without rotation of the coordinate axes.

The transformed equation of the line: $5x-4y-13=0$ is determined by replacing old coordinates by setting $x=X+(-2)=X-2$ & $y=Y+(-2)=Y-2$ in the equation of line, we get $$5(X-2)-4(Y-2)-13=0$$$$\implies \color{#0b4}{ 5X-4Y=15}$$ Where, $X$ & $Y$ are the new coordinates for the new origin $(-2, -2)$. The answer is correct which can be easily checked by plotting the given line & shifting the origin.

You notice that there is no option for this answer which actually leads to the fact that there is probably some error in the options provided because you may check that the answer obtained is correct just by plotting even rough plotting.