Answer :

Formula used:


(5, 9), (5, 16) and (29, 9)


Let the points be A (5, 16), B (5, 9) and C (29, 9)


Distance of AB


AB = √ ((5 – 5)2 + (9 – 16)2)


AB = √ ((0)2 + (–7)2)


AB = √ (0 + 49)


AB = √49


Distance of BC


B C= √ ((29 – 5)2 + (9 – 9)2)


BC = √ ((24)2 + (0)2)


BC = √ (576 + 0)


BC = √576


Distance of AC


AC = √ ((29 – 5)2 + (9 – 16))2)


AC = √ ((24)2 + (–7)2)


AC = √ (576 + 49)


AC = √ 625


i.e. AB2 + BC2


= (√49)2 + (√576)2


= 49 + 576


= 625 = (AC)2


Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Find the distanceTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I

Show that the folTamilnadu Board Math Term-I