Question

Which statement proves that △XYZ is an isosceles right triangle? XZ ⊥ XY XZ = XY = 5 The slope of XZ is , the slope of XY is , and XZ = XY = 5. The slope of XZ is , the slope of XY is , and the slope of ZY = 7.

Answer 1

Answer:

The slope of XZ is 3/4 , the slope of XY is -4/3 , and XZ = XY = 5 ⇒ 3rd answer

Step-by-step explanation:

* Lets look to the attached figure to solve the problem

- To prove that the Δ XYZ is an isosceles right triangle, you must

  find two sides the product of their slopes is -1 and they are equal

  in lengths

- From the figure the vertices of the triangle are;

  X = (1 , 3) , Y = (4 , -1) , Z = (5 , 6)

- The slope of the line whose endpoints are (x1 , y1) and (x2 , y2)

  is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

∵ The slope of $XY=\frac{-1-3}{4-1}=\frac{-4}{3}$

∵ The slope of $XZ=\frac{6-3}{5-1}=\frac{3}{4}$

∴ The slope of XY = -4/3 , the slope of XZ = 3/4

∵ -4/3 × 3/4 = -1

∴ XY ⊥ XZ

∴ ∠ X is a right angle

∴ Δ XYZ is a right triangle

- The distance between the two points (x1 , y1) and (x2 , y2) is

  $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

∵ $XY=\sqrt{(4-1)^{2}+(-1-3)^{2}}=\sqrt{9+16}=\sqrt{25}=5$

∵ $XZ=\sqrt{(5-1)^{2}+(6-3)^{2}}=\sqrt{16+9}=\sqrt{25}=5$

∴ XY = XZ = 5

∴ Δ XYZ is an isosceles right triangle

* The statement which prove that is:

 The slope of XZ is 3/4 , the slope of XY is -4/3 , and XZ = XY = 5