Answer :

Formula used:


Let the point A (–5, 2), B (9, –2) and C be the point on y–axis i.e. (0, y)


Distance of AC


AC = √ ((0 – (–5))2 + (y – 2)2)


AC = √ ((0 + 5)2 + (y – 2)2)


AC = √ ((5)2 + (y – 2)2)


AC = √ (25 + y2 – 4y + 4)


AC = √ y2 – 4y + 29


Distance of BC


BC = √ ((0 – 9)2 + (y – (–2)2)


BC = √ ((0 – 9)2 + (y + 2)2)


BC = √ ((9)2 + (y + 2)2)


BC = √ (81 + y2 + 4y + 4)


BC = √ y2 + 4y + 85


i.e. AC = BC ( Given)


√ y2 – 4y + 29 = √ y2 + 4y + 85


Squaring both sides


y2 – 4y + 29 = y2 + 4y + 8


y2 – 4y + 29 – y2 – 4y – 85 = 0


–8y – 56 = 0


–8 (y + 7) = 0


y + 7 = 0


y = –7


the point on y–axis is (0, –7).


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