Question

Show that the lines x-7y-22=0,3x+4y+9=0and 7x+y-54=0 form a right angled triangle

Answer 1

Answer:

x-7y-22=0 -------------------(1)

3x+4y+9=0 -----------------(2)

7x+y-54=0 ------------------(3)

Let A,B and C are the angles between the lines (1),(2); (2),(3) and (3),(1) respectively.

∴, cosA

=[|1×3+(-7)×4|]/√[1²+(-7)²×√3²+4²]  [∵, cosθ=|a₁a₂+b₁b₂|/(√a₁²+b₁²)×(√a₂²+b₂²)]

=|3-28|/√50√25

=25/(5√50)

=5/√50

=5/5√2

=1/√2

cosA=1/√2

or, cosA=cos45°

or, A=45°

cosB

=|3×7+4×1|/[(√3²+4²)×(√7²+1²)]

=|21+4|/(√25×√50)

=25/(5×5√2)

=1/√2

∴, B=45°

cosC

=|7×1+(-7)×1|/[(√7²+1²)×{√1²+(-7)²}]

=|7-7|/(√50×√50)

=0

cosC=0

or, cosC=cos90°

or, C=90°

Since, A=B and C=90° therefore the given lines form a right-angeled isosceles triangle.

Step-by-step explanation: