Question

question(+10) how do we know if a figure is an isosceles triangle? you have to check whether it has two congruent sides and angles. part i show all the necessary steps! for problems 1 - 6, find the distance between each pair of points. round to nearest tenth when necessary. 1. (0,0) and (-3,4) 2. (1,2) and (6,14) 3. (8,11) and (15,35) 4. (3,8) and (-5,-7) 5. (-2,-3) and (-1,-4) 6. (-3,5) and (3,5) 7. the points a(0,0), b(3,4), and c(-1,1) form a triangle. is it an isosceles triangle? justify your answer.

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Solve problem 1

For points $(0,0)$ and $(-3,4)$, substitute $x_1 = 0,y_1 = 0,x_2=-3,y_2 = 4$ into the formula:
\[

$$\begin{align*} d&=\sqrt{(-3 - 0)^2+(4 - 0)^2}\\ &=\sqrt{(-3)^2+4^2}\\ &=\sqrt{9 + 16}\\ &=\sqrt{25}\\ &= 5 \end{align*}$$

\]

Step3: Solve problem 2

For points $(1,2)$ and $(6,14)$, substitute $x_1 = 1,y_1 = 2,x_2=6,y_2 = 14$ into the formula:
\[

$$\begin{align*} d&=\sqrt{(6 - 1)^2+(14 - 2)^2}\\ &=\sqrt{5^2+12^2}\\ &=\sqrt{25+144}\\ &=\sqrt{169}\\ &=13 \end{align*}$$

\]

Step4: Solve problem 3

For points $(8,11)$ and $(15,35)$, substitute $x_1 = 8,y_1 = 11,x_2=15,y_2 = 35$ into the formula:
\[

$$\begin{align*} d&=\sqrt{(15 - 8)^2+(35 - 11)^2}\\ &=\sqrt{7^2+24^2}\\ &=\sqrt{49+576}\\ &=\sqrt{625}\\ &=25 \end{align*}$$

\]

Step5: Solve problem 4

For points $(3,8)$ and $(-5,-7)$, substitute $x_1 = 3,y_1 = 8,x_2=-5,y_2 = -7$ into the formula:
\[

$$\begin{align*} d&=\sqrt{(-5 - 3)^2+(-7 - 8)^2}\\ &=\sqrt{(-8)^2+(-15)^2}\\ &=\sqrt{64 + 225}\\ &=\sqrt{289}\\ &=17 \end{align*}$$

\]

Step6: Solve problem 5

For points $(-2,-3)$ and $(-1,-4)$, substitute $x_1=-2,y_1 = -3,x_2=-1,y_2 = -4$ into the formula:
\[

$$\begin{align*} d&=\sqrt{(-1-(-2))^2+(-4 - (-3))^2}\\ &=\sqrt{(-1 + 2)^2+(-4+3)^2}\\ &=\sqrt{1^2+(-1)^2}\\ &=\sqrt{1 + 1}\\ &=\sqrt{2}\approx1.4 \end{align*}$$

\]

Step7: Solve problem 6

For points $(-3,5)$ and $(3,5)$, substitute $x_1=-3,y_1 = 5,x_2=3,y_2 = 5$ into the formula:
\[

$$\begin{align*} d&=\sqrt{(3-(-3))^2+(5 - 5)^2}\\ &=\sqrt{(3 + 3)^2+0^2}\\ &=\sqrt{6^2}\\ &=6 \end{align*}$$

\]

Step8: Solve problem 7

Find the lengths of the sides of the triangle.
For side $AB$ with $A(0,0)$ and $B(3,4)$:
\[

$$\begin{align*} AB&=\sqrt{(3 - 0)^2+(4 - 0)^2}\\ &=\sqrt{3^2+4^2}\\ &=\sqrt{9 + 16}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]
For side $AC$ with $A(0,0)$ and $C(-1,1)$:
\[

$$\begin{align*} AC&=\sqrt{(-1 - 0)^2+(1 - 0)^2}\\ &=\sqrt{(-1)^2+1^2}\\ &=\sqrt{1+1}\\ &=\sqrt{2}\approx1.4 \end{align*}$$

\]
For side $BC$ with $B(3,4)$ and $C(-1,1)$:
\[

$$\begin{align*} BC&=\sqrt{(-1 - 3)^2+(1 - 4)^2}\\ &=\sqrt{(-4)^2+(-3)^2}\\ &=\sqrt{16 + 9}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]
Since $AB = BC=5$, the triangle $ABC$ is an isosceles triangle.

Answer:

1. $5$
2. $13$
3. $25$
4. $17$
5. $\approx1.4$
6. $6$
7. It is an isosceles triangle.