Without using the distance formula show that the points A(4,-2),B(-4,4) and C(10,6) are the vertices of a right angled triangle
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81
slope=(y2-y1)/(x2-x1)
slope of AB=(4-(-2))/(-4-4)=-(6/8)=-(3/4)
slope of BC=(6-(4))/(10-(-4))=(2/14)=(1/7)
slope of CA=(-2-6)/(4-10)=(8/6)=(4/3)
slope of AB * slope of CA=-(3/4)*(4/3)=-1
so AB is perpendicular to CA
thus,ABC are the vertices of a right angled triangle.
Answered by
17
Answer:slope=(y2-y1)/(x2-x1)
slope of AB=(4-(-2))/(-4-4)=-(6/8)=-(3/4)
slope of BC=(6-(4))/(10-(-4))=(2/14)=(1/7)
slope of CA=(-2-6)/(4-10)=(8/6)=(4/3)
slope of AB * slope of CA=-(3/4)*(4/3)=-1
so AB is perpendicular to CA
thus,ABC are the vertices of a right angled triangle.
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