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)

⇒ *y*^{2} – 4*y* + 3 = –1 (*x*^{2} + 3*x* – 4)

⇒ *y*^{2} – 4*y* + 3 = –*x*^{2} – 3*x* + 4

⇒ *x*^{2} + *y*^{2} + 3*x* – 4*y*= 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!!

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