Respuesta :

  • Slope of AC×Slope of BC=-1

[tex]\\ \rm\hookrightarrow \dfrac{7-1}{7-1}\times \dfrac{7-4}{7-10}=-1[/tex]

[tex]\\ \rm\hookrightarrow \dfrac{6}{6}\times \dfrac{3}{-3}=-1[/tex]

[tex]\\ \rm\hookrightarrow 1(-1)=-1[/tex]

Hence they are vertices of right angled triangle

C is the right angle

semsee45

Answer:

Determine the slopes (gradients) of each line.

If the slopes of two of the lines are negative reciprocals (if their product is -1), the lines are perpendicular (the vertex is 90°).

Using slope formula: [tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

Slope of AC:

A = (1, 1) = [tex](x_1,y_1)[/tex]

C = (7, 7) = [tex](x_2,y_2)[/tex]

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{7-1}{7-1}=1[/tex]

Slope of AB:

A = (1, 1) = [tex](x_1,y_1)[/tex]

B = (10, 4) = [tex](x_2,y_2)[/tex]

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-1}{10-1}=\dfrac13[/tex]

Slope of BC:

B = (10, 4) = [tex](x_1,y_1)[/tex]

C = (7, 7) = [tex](x_2,y_2)[/tex]

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{7-4}{7-10}=-1[/tex]

Therefore, as AC and BC are negative reciprocals of each other (1 x -1 = -1), m∠C = 90° which proves that the triangle is a right triangle.

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