Question

write the co ordinates of a point on x axis which is equidistant from the point
(-3, 4) ; (2, 5) ​

Answer 1

Answer :

The required co-ordinates of a point on X-axis is (2/5 , 0)

Given :

• The points are (-3 , 4) and (2 , 5)

To Find :

• The point on X-axis which is equidistant from the given points.

Formula to be used :

If (x₁ , y₁) and (x₂ , y₂) are two points then the distance between them is given by :

$\sf \longrightarrow \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Solution :

Let us consider the points A(-3 , 4) and B(2 , 5). Also let the point on X-axis be C(a , 0) which is equidistant from A and B

According to question :

AC = BC

$\sf \sqrt{ \{a-(-3)\}^{2} + ( 0- 4)^{2} } = \sqrt{(2-a)^{2} + (5-0)^{2} } \\\\ \sf \implies (a+3)^{2}+16 = 4 + a^{2} - 4a + 25 \\\\ \sf \implies a^{2} + 9 + 6a + 16 = 4 + a^{2} - 4a + 25 \\\\ \sf \implies 6a + 25 = 29 - 4a \\\\ \sf \implies 6a + 4a = 29 - 25 \\\\ \sf \implies 10a = 4 \\\\ \sf \implies a = \dfrac{4}{10}\\\\ \sf \implies a = \dfrac{2}{5}$

$\sf The \: \: coordinates \: \: is \: \: (\dfrac{2}{5} , 0)$

Answer 2

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$\huge\tt{TO~FIND:}$

• write the co ordinates of a point on x axis which is equidistant from the point (-3, 4) ; (2, 5)

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$\huge\tt{CONCEPT~USED:}$

• If (x1 , y1) and (x2 , y2) are two points then the distance between them is given by ➡√(x2- x1)²+ (y2 - y1)²

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$\huge\tt{SOLUTION:}$

Let us consider the points A(-3 , 4) and B(2 , 5). Also let the point on X-axis be C(a , 0) which is equidistant from A and B,

Now,

↪AC = BC

↪√{a - (-3)}² + (0-4)² = √(2 - a)² + (5-0)²

↪(a + 3)² + 16 = 4 + a² - 4a + 25

↪a² + 9 + 6a + 16 = 4 + a² - 4a + 25

↪6a + 25 = 29 - 4a

↪6a + 4a = 29 - 25

↪10a = 4

↪a = 4/10

↪a = 2/5

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