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 $(-2,-2)$].

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


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

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

I look for short/simple method.

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

  • $\begingroup$ In every Cartesian coordinate system familiar to me, the origin lies at $(0,0)$--in that system. If someone wants to say the origin of a coordinate system lies at $(-2,-2)$ in some other system, they really should say which system's origin lies at $(-2,-2)$ in which system. Is the origin of the first system at $(-2,-2)$ in the second system, or the other way around? And did they want the equation of the perpendicular line in the old system or in the new system? $\endgroup$ – David K May 27 '15 at 1:53

$$ 5x-4y=13 $$ 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 $y$, $$ 5(x+2)-4(y+2)=13 $$ $$ 5x-4y=13-2 $$ $$ 5x-4y=11 $$


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.

  • $\begingroup$ If the origin of the new coordinate system has origin $(-2, -2)$ in the old coordinate system, then in the new coordinate system, the old origin is at the point $(2, 2)$, so $x = X + 2$ and $y = Y + 2$. $\endgroup$ – N. F. Taussig May 27 '15 at 2:14
  • $\begingroup$ The coordinate axes are shifted from old origin to a new origin but not from the new to the old one. You should see the new origin $(-2, -2)$ in the perspective of the old one $(0, 0)$. $\endgroup$ – Harish Chandra Rajpoot May 27 '15 at 2:22
  • $\begingroup$ OK! you may better understand by a rough plot. Have you tried it by a rough diagram even? you may then realize the fact. $\endgroup$ – Harish Chandra Rajpoot May 27 '15 at 2:24
  • $\begingroup$ Your work is correct. To obtain the answer in the book, you have to express the equation of the line as $5(x + 2) - 4(y + 2) = 13$, which yields $5x - 4y = 11$ in terms of the variables in the old coordinate system. In the new coordinate system, the point $(1, -2)$ in the old coordinate system becomes the point $(3, 0)$, so we obtain $y = \frac{5}{4}(X - 3) \Rightarrow 5x - 4y = 15$, as you said. Since $X = x + 2$ and $Y = y + 2$, I reversed the roles of the variables when I read the question late at night. $\endgroup$ – N. F. Taussig May 27 '15 at 11:32

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