Suppose I have a point $P(x_1, y_1$) and a line $ax + by + c = 0$. I draw a perpendicular from the point $P$ to the line. The perpendicular meets the line at point $Q(x_2, y_2)$. I want to find the coordinates of the point $Q$, i.e., $x_2$ and $y_2$.

I searched up for similar questions where the coordinates of end points of the line segment are given. But here, I've got an equation for the line. So, I am pretty clueless how to solve this.

Please give me a formula to arrive at my answer (if any) and show me its derivation too. I am a high school student with a basic knowledge of trigonometry. I have no idea of calculus, so please give me a simplified answer.

Any help is highly appreciated. Thanks a lot in advance...

  • Choose any other point in the line say R and then the vectors PQ and QR are perpendicular so their dot product will give 0, if you are familiar with vectors – arberavdullahu Oct 29 '16 at 8:09
up vote 0 down vote accepted

As the line $PQ$ is perpendicular to the other line, its equation is

$$bx-ay+d=0.$$

Expressing that it contains $(x_1,y_1)$, you get

$$bx-ay=bx_1-ay_1.$$

Now solve the system

$$\begin{cases}bx_2-ay_2=bx_1-ay_1,\\ax_2+by_2=-c.\end{cases}$$

  • Thank you so much for the answer. By the way, can you please tell me or direct me as to how PQ got its equation... I know that the slope of the perpendicular line is the negative inverse of that of the equation of the given line. But, how did you get bx − ay+ d = 0? – Hariharan Nov 2 '16 at 18:35
  • @Hariharan: write the slopes of the two lines. – Yves Daoust Nov 2 '16 at 19:00

Use a parametric representation of this perpendicular: it is directed by the vector $(a,b)$ since the latter is normal to the given line. Thus a parametric representation is $$(x,y)=(x_1,y_1)+t(a,b),\quad t\in\mathbf R.$$ A point on this perpendicular is on the given line if it satisfies its equation: $$a(x_1+ta)+b(y_1+tb)+c=0,$$ whence the corresponding value of $t$.

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