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Q7:
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Solution :
Given: Two points $A(2, -5)$ and $B(-2, 9)$. A point $P$ lies on the $x$-axis such that it is equidistant from $A$ and $B$.
To Find: The coordinates of point $P$.
Step 1: Defining the coordinates of point P
Since point $P$ lies on the $x$-axis, its $y$-coordinate must be $0$. Let the coordinates of point $P$ be $(x, 0)$.
Step 2: Applying the Distance Formula
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Step 3: Setting up the Equidistance Equation
Since $P$ is equidistant from $A(2, -5)$ and $B(-2, 9)$, we have $PA = PB$.
Therefore, $PA^2 = PB^2$ [Squaring both sides to eliminate the square root].
Step 4: Calculating the squared distances
For $PA^2$:
$PA^2 = (x - 2)^2 + (0 - (-5))^2$
$PA^2 = (x - 2)^2 + (5)^2$
$PA^2 = x^2 - 4x + 4 + 25$
$PA^2 = x^2 - 4x + 29$
For $PB^2$:
$PB^2 = (x - (-2))^2 + (0 - 9)^2$
$PB^2 = (x + 2)^2 + (-9)^2$
$PB^2 = x^2 + 4x + 4 + 81$
$PB^2 = x^2 + 4x + 85$
Step 5: Solving for x
Equating $PA^2 = PB^2$:
$x^2 - 4x + 29 = x^2 + 4x + 85$
Subtract $x^2$ from both sides:
$-4x + 29 = 4x + 85$
Subtract $4x$ from both sides:
$-8x + 29 = 85$
Subtract $29$ from both sides:
$-8x = 85 - 29$
$-8x = 56$
Divide by $-8$:
$x = \frac{56}{-8}$
$x = -7$
Final Answer: The point on the x-axis equidistant from (2, –5) and (–2, 9) is (-7, 0).
More Questions from Class 10 Mathematics Coordinate geometry EXERCISE 7.1
- Q1(i): Find the distance between the following pairs of points : (i) (2, 3), (4, 1)
- Q1(ii): Find the distance between the following pairs of points : (ii) (– 5, 7), (– 1, 3)
- Q1(iii): Find the distance between the following pairs of points : (iii) ($a$, $b$), (– $a$, – $b$)
- Q10: Find a relation between $x$ and $y$ such that the point ($x$, $y$) is equidistant from the point (3, 6) and (– 3, 4).
- Q2: Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B.
- Q3: Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
- Q4: Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
- Q5: In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
- Q6(i): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
- Q6(ii): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
- Q6(iii): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (iii) (4, 5), (7, 6), (4, 3), (1, 2)
- Q8: Find the values of $y$ for which the distance between the points P(2, – 3) and Q(10, $y$) is 10 units.
- Q9: If Q(0, 1) is equidistant from P(5, –3) and R($x$, 6), find the values of $x$. Also find the distances QR and PR.
CBSE Solutions for Class 10 Mathematics Coordinate geometry
Chapters in CBSE - Class 10 Mathematics
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