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!!

  • 31

 yes plz explain the line

It can be clearly seen that (α, β) = (1, 1) satisfies the equation (1).

plz

  • 10
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