Question

Victor draws one side of equilateral ∆PQR on the coordinate plane at points P (–9, –2) and Q (–2, –2). What are the two possible coordinates of vertex R? Round to the nearest tenth.

Answer 1

Answer:
`(-5.5,4.062)\ \text{and}\ (-5.5,-8.062)`

Explanation:

Given

The two vertices of a equilateral triangle are
`(-9,-2)\ \text{and}\ (-2,-2)`

Suppose the third vertex is
`(x,y)`

Side length in equilateral triangle are equal.

`\Rightarrow √(\left( -2+9\right)^2+\left( -2+2\right)^2)=√(\left( x+9\right)^2+\left( y+2\right)^2)\\\text{squaring both sides}\\\\\Rightarrow 7^2=\left( x+9\right)^2+\left( y+2\right)^2\quad \ldots(i)\\\\`

Similarly,

`\Rightarrow √(\left( -2+9\right)^2+\left( -2+2\right)^2)=√(\left( x+2\right)^2+\left( y+2\right)^2)\\\\\Rightarrow 7^2=\left( x+2\right)^2+\left( y+2\right)^2\quad \ldots(ii)`

Subtract (i) and (ii)

`\Rightarrow \left( x+9\right)^2-\left(x+2\right)^2=7^2-7^2\\\\\Rightarrow (x+9+x+2)(x+9-x-2)=0\\\Rightarrow (2x+11)7=0\\\\\Rightarrow x=-(11)/(2)`

Insert the value of
`x` in equation (ii)

`\Rightarrow \left(-5.5+2\right)^2+\left(y+2\right)^2=7^2\\\\\Rightarrow \left(-5.5+2\right)^2+\left(y+2\right)^2-\left(-5.5+2\right)^2=7^2-\left(-5.5+2\right)^2\\\\\Rightarrow \left(y+2\right)^2=36.75\\\\\Rightarrow y=4.06217\dots ,\:y=-8.06217`

So, two possible vertex are
`(-5.5,4.062)\ \text{and}\ (-5.5,-8.062)`