Question

show that the point (- 2 5) ,(3 - 4l and (7,10) are the vertices of the right angle triangle of isosceles triangle​

Answer 1

Answer:

Step-by-step explanation:

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

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

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Let A(3, -4), B(7, 10) and C(- 2, 5) be the vertices of Δ ABC.

AB = √[(3 - 7)² + (- 4 - 10)²] = √212 = 2√53

AC = √[(- 2 - 3)² + (5 + 4)²] = √106

BC = √[(- 2 - 7)² + (5 - 10)²] = √106

So far, the triangle is isosceles AC = BC.

Now we have to show, that triangle is right angled.

Let find the slope of the line passing through the points A(3, - 4) and C(- 2, 5) , and the slope of the line passing through the points B(7, 10) and C(- 2, 5)

$m_{AC}$ = (5 + 4) / (- 2 - 3) = - $\frac{9}{5}$

$m_{BC}$ = (5 - 10) / (- 2 - 7) = $\frac{5}{9}$

As we can see $m_{AC} = - \frac{1}{m_{BC} }$  ⇒ AC ⊥ BC ⇒ m∠ACB = 90°

Thus, Δ ABC is the isosceles right angled triangle.