Answer:
Refer picture enclosed
Step-by-step explanation:
Given is a line BA and a point P. We have to construct a line perpendicular to BA through P.
Case I: P lies on the line BA
Without loss of generality assume BP is smaller than AP.
Cut an arc C on AB from P such that BP=CP
Now with B and C as centres draw arc up and down the line with more than half length of BC
Join the intersecting of arcs. That line is the perpendicular for BA through P.
This is because of concurrency of two triangles above the line BC
Case2:
P lies outside the line. Measure PB and cut an arc of BA with length PB
Let the point be C
Then CP=BP
NOw do as we did before the line with arcs more than 1/2 length of BC from B and C only down. Join the point of intersection down with P. This line will be perpendicular through P to BA
Answer: place the compass on point A. Open the compass to a point somewhere between point P and point B
ON GRADPOINT.
