If two lines intersect in such a way that they make a right angle at the point of intersection. Then, they are perpendicular to each other.

**Perpendicular to a line if a point is given on it (using a ruler and a set square)**

Step 1: A line l is given and point P on it.

Step 2: Put the ruler with one of its next to l.

Step 3: Put the set square with one of its sides by the side of the already aligned edge of the ruler. So, the right-angled corner comes in contact with the ruler.

Step 4: Slide the set-square so that the right-angled corner coincides with P.

Step 5: Hold the set-square. Draw PQ along the edge of the set-square.

PQ is the required perpendicular to l from the given point P.

**Perpendicular to a line if a point is given on it (using a ruler and compass)**

Draw a perpendicular of the line at a given point A.

Step 1: Take A as the centre and draw a big arc with any radius. So, it intersects the line at point B and C.

Step 2: Take B as the centre and draw an arc with the radius more than AC and then take C as centre and draw an arc. So that, they intersect each other at point D

Step 3: Join AD. AD is perpendicular bisector to CB.

**Perpendicular to a line from a point which is not on it (using ruler and set square)**

Step 1: P is the point outside the given line l.

Step 2: Put the set-square on l so that one side of its right angle comes on l.

Step 3: Put a ruler on the hypothenuse of set-square.

Step 4: Hold the ruler and slide the set-square until it touches the point P.

Step 5: Join PM.

Now

**The perpendicular bisector of a line-segment (using ruler and compass)**

Step 1: Draw a line-segment AB.

Step 2: Take A as the centre and draw two arcs- one upside and one downside with the radius of more than half of the length of AB. Or, you can draw a circle taking A as the centre for the convenience.

Again, make the arcs with taking B as the centre so that they intersect the previous arcs.

Step 3: Join the intersections of the arcs and name them as C and D.

Step 4: The required perpendicular bisector of AB is CD. Hence, AO = OB.

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