MATHS 1 Marks Question

Proof : Here we are given two lines l and m. We need to prove that they have only one point in common.
For the time being, let us suppose that the two lines intersect in two distinct points,
say P and Q. So, you have two lines passing through two distinct points P and Q. But
this assumption clashes with the axiom that only one line can pass through two distinct
points. So, the assumption that we started with, that two lines can pass through two
distinct points is wrong.
From this, what can we conclude? We are forced to conclude that two distinct
lines cannot have more than one point in common.

Euclid's Axiom 5 states that  "The whole is greater than the part. Since this is true for anything in any part of the world. So, this is a universal truth.

Let a line AB have two mid-points, say, C and D. Then
AB = AC + CB = 2AC . . . . (i) . . . [As C is the mid-point of AB]
and AB = AD + DB = 2AD . . . . (ii) [As D is the mid-point of AB]
From equation (i) and (ii)
AC = AD and CB = DB
But this will possible only when D lies on point C. So every line segment has one and only one mid-point.

The length of the line-segment joining the centre of a circle to any point on its circumference is called the raduis of the circle.

A line segment is a part of a line with two end points and it cannot be extended further. It has a definite length or breadth.

Two lines which are at a right angles to each other are called perpendicular lines.

Lines which do not intersect anywhere are called parallel lines.

Take any line l and a point P not on l. Therefore, by Playfair’s axiom, which is equivalent to the fifth postulate, which states that there is a unique line m through P which is parallel to l.
Now, the distance of a point from a line is the length of the perpendicular from the point to the line. This distance will be the same for any point on m from l and any point on l from m. Therefore, these two lines are everywhere equidistant from one another.