Question

prove that through a given point, we can only draw one perpendicular to a given line.

Answer 1

Answer:

[Hint : Use proof by contradiction].

Step-by-step explanation:

From the point P , a perpendicular PM is drawn to the given line AB.

∴ ∠ PMB = 90

∘

Let if possible , we can draw another perpendicular PN to the line AB. Then ,

∠ PMB = 90

∘

∴ ∠ PMB = ∠ PNB , which is possible only when PM and PN coincide with each other.

Hence , through a given point , we can draw only one perpendicular to a given line.

solution

Answer 2

Answer:

Lets draw a line $AB$ and mark a point $P$ such that $P$ does not lie on $AB$

Now lets draw a line $PM$ perpendicular to $AB$ from point $P$.

Lets say we can draw another perpendicular line $PN$ to $AB$ through point $P$

We know that the shortest distance of a point from a line is its perpendicular distance.

In other words, we can say that the shortest distance of a point from a line is a constant.

⇒ $PN = PM = k$

Applying the Pythagoras Theorem

   $PM^{2} + MN^{2} = PN^{2}$

We know that $PM = PN = k$

⇒ $k^{2} +MN^{2} = k^{2}$

⇒ $MN^{2} = k^{2} - k^{2}$

⇒ $MN = 0$

Since, $MN = 0$ this means that the distance between the lines is zero.

This means that $PM$ and $PN$ coincide with each other.

∴ Through a given point, we can only draw one perpendicular to a given line.