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Question

Find the area of the triangle formed by the straight lines whose equations are
y=m1x+am1, y=m2x+am2, and y=m3x+am3.

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Solution

y=m1x+am1........(i)y=m2x+am2........(ii)y=m3x+am3........(iii)

solving (i) and (ii) by y from (ii) in (i)

m1x+am1=m2x+am2(m2m1)x=am1am2(m2m1)x=a(1m11m2)=am2m1m1m2x=am1m2y=m1×1m1m2+am1y=a(1m2+1m1)

So the point of intersection is A(am1m2,a(1m1+1m2))

solving (ii) and (iii)

B(am2m3,a(1m2+1m3))

solving (iii) and (i)

C(am3m1,a(1m3+1m1))

So the vertices of triangle are A(am1m2,a(1m1+1m2)),B(am2m3,a(1m2+1m3)),C(am3m1,a(1m3+1m1))

Area of =12∣ ∣ ∣ ∣ ∣ ∣ ∣am1m2a(1m1+1m2)1am2m3a(1m2+1m3)1am3m1a(1m3+1m1)1∣ ∣ ∣ ∣ ∣ ∣ ∣

Simplifying using properties of determinant

=12a2∣ ∣ ∣ ∣ ∣ ∣1m1m21m1+1m211m2m31m2+1m311m3m11m3+1m11∣ ∣ ∣ ∣ ∣ ∣=12a2∣ ∣ ∣ ∣ ∣ ∣1m1m21m2m31m1+1m21m21m311m2m31m3m11m2+1m31m31m111m3m11m3+1m11∣ ∣ ∣ ∣ ∣ ∣

=12a2∣ ∣ ∣ ∣ ∣ ∣m3m1m1m2m3m3m1m3m10m1m2m1m2m3m1m2m2m101m3m11m3+1m11∣ ∣ ∣ ∣ ∣ ∣

Expanding along C3

=12a2{0+0+1(m3m1m1m2m3×m1m2m2m1m3m1m3m1×m1m2m1m2m3)}=12a2(m3m1)(m1m2)(1m21m22m31m21m23m2)=12a2(m3m1)(m1m2)(m3m2m21m22m23)=a2(m3m1)(m1m2)(m3m2)÷2m21m22m23

Hence proved.


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