The points P (x1, y1) and Q (x2 , y2) lie on the same side or on opposite sides of the line a x +b y +c = 0 according as the expressions a x1 + b y1 +c and a x2 + by2 +c have same sign or opposite signs.
The sides of a triangle are given by the equations 3 x +4 y =
10, 4 x -3 y = 5 and 7 x +y +10 = 0. Show that the origin lies with in
the triangle.
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The given lines are
3 x +4 y -10 = 0 ...(i)
4 x -3 y -5 = 0 ...(ii)
and 7 x + y +10 = 0 ...(iii)
Let ABC be the triangle formed by these lines.
Solving these equations simultaneously, taking two at a time, the vertices of the triangle are
A (-2, 4), B (2, 1) and C (-1, -3)
On substituting (-2, 4) in L.H.S. of (ii), we get -8 -12 -5 = -25
and substituting (0, 0) in L.H.S. of (ii), we get 0 -0 -5 = -5
Since both have same sign, therefore, origin and A lie on the same side of BC.
Similarly origin and B lie on the same side of CA; and origin and C lie on the
same side of AB. (Please check it)
From these results, it follows that the origin lies with in the triangle
formed by the given lines.