Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units
Solution:
Given, the distance between the points A(-3, -14) and B(a, -5) is 9 units.
We have to find the value of a.
The distance between two points P (x₁ , y₁) and Q (x₂ , y₂) is
√[(x₂ - x₁)² + (y₂ - y₁)²]
Now, distance between the points A (-3, -14) and B(a, -5) is
√[(a - (-3))² + (-5 - (-14))²] = 9
√[(a + 3)² + (-5 + 14)²] = 9
√[(a + 3)² + (9)²] = 9
On squaring both sides,
(a + 3)² + 81 = 81
(a + 3)² = 81 - 81
(a + 3)² = 0
Taking square root,
a + 3 = 0
a = -3
Therefore, the value of a is -3.
✦ Try This: The centre of a circle is (2a, a - 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7
NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 4
Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units
Summary:
The value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units is -3
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