✏️DISTANCES
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Problem: What type of triangle is formed by the vertices P(-2,1), Q(3,2), and R(0,-2)?
Solution: Find the distances between all of these points connected indicating as the sides of the triangle using the distance formula.
[tex]\begin{aligned} & \bold{\color{lightblue}Formula:} \\&\boxed{d =\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2\,}} \end{aligned}[/tex]
- Find side PQ.
- [tex]\begin{aligned}{PQ = \sqrt{(3-( \text - 2))^2 + (2 - 1)^2\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{PQ = \sqrt{(3 + 2)^2 + (2 - 1)^2\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{PQ = \sqrt{5^2 + 1^2\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{PQ = \sqrt{25 +1\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{PQ = \sqrt{26\,}} \end{aligned}[/tex]
- Find side QR.
- [tex]\begin{aligned}{QR = \sqrt{(0 - 3)^2 + ( \text - 2 - 2)^2\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{QR = \sqrt{( \text- 3)^2 + ( \text -4)^2\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{QR = \sqrt{9 + 16\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{QR = \sqrt{25\,}} \end{aligned}[/tex]
- Find side RP
- [tex]\begin{aligned}{RP = \sqrt{( \text - 2 - 0)^2 + (1 - (\text - 2) )^2\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{RP = \sqrt{( \text - 2 - 0)^2 + (1 + 2 )^2\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{RP = \sqrt{( \text - 2)^2 + (3)^2\,}} \end{aligned}[/tex]
- [tex]\begin{aligned}{RP = \sqrt{4 + 9\,}} \end{aligned}[/tex]
- [tex]RP = \sqrt{13} [/tex]
- Since, all the sides have different lengths, this kind of triangle is:
- [tex] \large \rm Kind \: of \: Triangle = \boxed{ \rm \green{ \: Scalene \: }}[/tex]
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