Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendi

1.	Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.
Answer» Let m₁ be the slope of the line x – 2y – 7 = 0 ∴ m₁ = \(\frac {-coefficient\, of\, x}{coefficient\, of\, y} = \frac {-1}{2}= \frac {1}{2}\) Let m₂ be the slope of the line 2x + y + 1 = 0. ∴ m₂ = \(\frac {-coefficient\, of\, x}{coefficient\, of\, y} = \frac {-2}{1}= -2\) Since m x m = 1/2 x (-2) = -1, the given lines are perpendicular to each other. Consider x – 2y – 7 = 0 …(i) 2x + y + 1 =0 …(ii) Multiplying equation (ii) by 2, we get 4x + 2y + 2 = 0 …(iii) Adding equations (i) and (iii), we get 5x – 5 = 0 ∴ x = 1 Substituting x = 1 in equation (ii), we get 2 + y + 1 = 0 ∴ y = – 3 ∴ The point of intersection of the given lines is (1,-3).

Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.

Answer»

Let m₁ be the slope of the line x – 2y – 7 = 0

∴ m₁ = \(\frac {-coefficient\, of\, x}{coefficient\, of\, y} = \frac {-1}{2}= \frac {1}{2}\)

Let m₂ be the slope of the line 2x + y + 1 = 0.

∴ m₂ = \(\frac {-coefficient\, of\, x}{coefficient\, of\, y} = \frac {-2}{1}= -2\)

Since m x m = 1/2 x (-2) = -1,

the given lines are perpendicular to each other. Consider

x – 2y – 7 = 0 …(i)

2x + y + 1 =0 …(ii)

Multiplying equation (ii) by 2, we get 4x + 2y + 2 = 0 …(iii)

Adding equations (i) and (iii), we get 5x – 5 = 0 ∴ x = 1

Substituting x = 1 in equation (ii), we get 2 + y + 1 = 0

∴ y = – 3

∴ The point of intersection of the given lines is (1,-3).