Find a relation between x and y such that the point P(x, y) is equidistant from the
points A(5, 3) and B(1, 7)
Given
P(x, y) is equidistant from A and B,
PA = PB
squaring both sides
PA² = PB²
Using distance formula to find PA and PB
(x - 5)² + (y - 3)² = (x - 1)² + (y - 7
Expand both sides using identity
x² - 10x + 25 + y² - 6y + 9 = x² - 2x + 1 + y² - 14y + 49
-10x - 6y + 34 = -2x - 14y + 50
-10x + 2x - 6y + 14y = 50 - 34
-8x + 8y = 16
or
-x + y = 2
or
y = x + 2
watch this video for more details
cbse 10th maths 2025 2026 old board exam question paper coordinate geometry distance formula
No comments:
Post a Comment
please leave your comments