Question

Find the area of a triangle :$y=x,y=2x$ and $y=3x+4$ ?

Answer 1

First, we will find out the co-ordinates of the triangle’s three vertices. These occur at the points that a pair of lines intersect.

Vertex1:Thisiswherethelines$y=x$and$y=2x$intersect.Thishappenswhen$x=2x\Rightarrow x=0\Rightarrow y=0.$Thusthepointis$(0,0).$

Vertex2:Thisiswherethelines$y=x$andy=3x+4intersect.Thishappenswhenx=3x+4⇒x=−2⇒y=−2.Thusthepointis(-2,-2).

Vertex3:Thisiswherethelinesy=2xand$y=3x+4$intersect.Thishappenswhen$2x=3x+4\Rightarrow x=-4\Rightarrow y=-8.$Thusthepointis$(-4,-8)$.

The circumcentre of a triangle is the point $(a,b)$ such that a circle of radius r centre on $(a,b)$ passes through the three vertices.

Every point on this circle satisfies the generic equation:${(x-a)}^2+{(y-b)}^2=r^2$.Expanding this equation,we have$x^2-2ax+a^2+y^2-2by+b^2=r^2.$

At vertex1,$x=0$ and $y=0$,thus we have Equation$A:a^2+b^2=r^2$

At vertex2,$x=-2$and$y=-2$,thus we have Equation $B:4+4a+a^2+4+4b+b^2=r^2$

Atvertex3,$x=-4$and$y=-8$,thuswehaveEquation$C:16+8a+a^2+64+16b+b^2=r^2$

Subtracting Equation A from Equation B,we have Equation $D:8+4a+4b=0$

Subtracting Equation B from Equation C,we have Equation $E:72+4a+12b=0$

Subtracting Equation D from Equation E,we have Equation $F:64+8b=0\Rightarrow b=-8$

Substituting this into Equation $D:8+4a-32=0\Rightarrow a=6$

So,thecircumcentreis$(6,-8).$

As a bonus,using Equation A,the radius of the circle is 10.