Question

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

Answer 1

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.

Answer 2

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Step-by-step explanation:

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