Question

Without using distance formula show that the points a(4,-2) , b(-4,4) and c(10,6) are the vertices of a right angled triangle

Answer 1

yes it is right angle triangle

Answer 2

Answer:

The points are A(4,-2) , B(-4,4) and C(10,6) are vertices of a right angled triangle.

Step-by-step explanation:

The given points are A(4,-2) , B(-4,4) and C(10,6).

Slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Using slope formula,

$m_{AB}=\frac{4-(-2)}{-4-4}=\frac{6}{-8}=-\frac{3}{4}$

$m_{BC}=\frac{6-4}{10-(-4)}=\frac{2}{14}=\frac{1}{7}$

$m_{AC}=\frac{6-(-2)}{10-4}=\frac{8}{6}=\frac{4}{3}$

Using above slopes it is clear that the slopes are difference , so the points are not collinear and these points are vertices of a triangle.

Product of slopes of two perpendicular lines is -1.

$m_{AB}\times m_{AC}=-\frac{3}{4}\times \frac{4}{3}=-1$

It is AB is perpendicular to AC.On of the angle is 90 degree.

Therefore the points are A(4,-2) , B(-4,4) and C(10,6) are vertices of a right angled triangle.