# Given points M(0, 10), N(5, 0) and P(15, 15) in DeltaMNP and points M(0, 10), Q(10, -10), and R(30, 20) in DeltaMQR, how do we find that the two triangles are similar?

Jan 7, 2017

Find all the sides and check whether they are proportional.

#### Explanation:

Given two points $A \left({x}_{1} , {y}_{1}\right)$ and $B \left({x}_{2} , {y}_{2}\right)$, the distance between the two $A B = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

As we have points $M \left(0 , 10\right)$, $N \left(5 , 0\right)$ and $P \left(15 , 15\right)$ in $\Delta M N P$

and points $M \left(0 , 10\right)$, $Q \left(10 , - 10\right)$, and $R \left(30 , 20\right)$ in $\Delta M Q R$

and we have in $\Delta M N P$

$M N = \sqrt{{\left(5 - 0\right)}^{2} + {\left(0 - 10\right)}^{2}} = \sqrt{25 + 100} = \sqrt{125} = 5 \sqrt{5}$

$N P = \sqrt{{\left(15 - 5\right)}^{2} + {\left(15 - 0\right)}^{2}} = \sqrt{100 + 225} = \sqrt{325} = 5 \sqrt{13}$

$M P = \sqrt{{\left(15 - 0\right)}^{2} + {\left(15 - 10\right)}^{2}} = \sqrt{225 + 25} = \sqrt{250} = 5 \sqrt{10}$

and in $\Delta M Q R$

$M Q = \sqrt{{\left(10 - 0\right)}^{2} + {\left(- 10 - 10\right)}^{2}} = \sqrt{100 + 400} = \sqrt{500} = 10 \sqrt{5}$

$Q R = \sqrt{{\left(30 - 10\right)}^{2} + {\left(20 - \left(- 10\right)\right)}^{2}} = \sqrt{400 + 900} = \sqrt{1300} = 10 \sqrt{13}$

$M R = \sqrt{{\left(30 - 0\right)}^{2} + {\left(20 - 10\right)}^{2}} = \sqrt{900 + 100} = \sqrt{1000} = 10 \sqrt{10}$

Therefore $\frac{M N}{M Q} = \frac{N P}{Q R} = \frac{M P}{M R} = \frac{1}{2}$

and as sides of two triangles are proportional, we have

$\Delta M N P \approx \Delta M Q R$