Q.17 of miscellaneous ex. of N cert book chapter 10 st. line
Dear Student
Let the coordinates of the point of intersection of the perpendicular sides of the triangle be (x,y).
AB ⊥ BC
∴ Slope of AB × slope of BC = –1
⇒ (y – 3) (y – 1) = –1 (x – 1) (x + 4)
⇒ y2 – 4y + 3 = –1 (x2 + 3x – 4)
⇒ y2 – 4y + 3 = –x2 – 3x + 4
⇒ x2 + y2 + 3x – 4y= 1
Let (α, β) be the point satisfying the equation (1).
∴ Equation of line segment AB is,
and, equation of line segment BC is,
It can be clearly seen that (α, β) = (1, 1) satisfies the equation (1).
∴ Equation of line segment AB passing through points (1, 3) and (1, 1) is given by,
⇒ x – 1 = 0
⇒ x = 1.
Equation of line segment BC passing through points (1, 1) and (–4, 1) is given by,
⇒ y – 1 = 0
⇒ y = 1
∴ Equations of legs of the triangle are x = 1 and y = 1.
Cheers!!