What are the coordinates of point q?

Solution:

Given, the point Q on the x-axis lies on the perpendicular bisector of the line segment joining the points A(-5, -2) and B(4, -2).

We have to find the type of triangle formed by the points Q, A and B.

The point Q lies on the x-axis means Q = (x, 0)

We know that any point lying on the perpendicular bisector of the line segment joining two points is equidistant from the two points.

The perpendicular bisector joining the two points are A(-5, -2) and B(4, -2)

To check if the point Q lies on the perpendicular bisector, then distance between QA must be equal to distance between QB

i.e.,QA = QB

The distance between two points P (x₁ , y₁) and Q (x₂ , y₂) is

√[(x₂ - x₁)² + (y₂ - y₁)²]

Distance between Q(x, 0) and A(-5, -2) = √[(-5 - x)² + (-2 - 0)²]

= √[(-5 - x)² + (-2)²]

= √[(-5 - x)² + 4]

Distance between Q(x, 0) and B(4, -2) = √[(4 - x)² + (-2 - 0)²]

= √[(4 - x)² + (-2)²]

= √[(4 - x)² + 4]

Now, √[(-5 - x)² + 4] = √[(4 - x)² + 4]

On squaring both sides,

(-5 - x)² + 4 = (4 - x)² + 4

Canceling out common terms,

(-5 - x)² = (4 - x)²

By using algebraic identity,

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

25 + 10x + x² = 16 - 8x + x²

By grouping,

x² - x² + 10x + 8x + 25 - 16 = 0

18x + 9 = 0

18x = -9

x = -9/18

x = -1/2

Therefore, the point Q is (-1/2, 0)

Distance between A(-5, -2) and B(4, -2) = √[(4 - (-5))² + (-2 - (-2))²]

= √[(4+5)² + (-2 + 2)²]

= √81

= 9

Distance between Q(-1/2, 0) and B(4, -2) = √[(4 + (1/2))² + 4]

= √[(4.5)² + 4]

= √24.25

Distance between Q(-1/2, 0) and A(-5, -2) = √[(-5 + 0.4)² + 4]

= √[(4.5)² + 4]

= √24.25

It is clear that QA = QB

Two sides of a triangle are equal.

Therefore, the points Q, A and B form an isosceles triangle.

✦ Try This: The perpendicular bisector of the line segment joining the points A(1, 5) and 8(4, 6) cuts the y-axis at

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 6

Summary:

The coordinates of the point Q on the x–axis which lies on the perpendicular bisector of the line segment joining the points A (–5, –2) and B(4, –2) is (-1/2, 0). The type of triangle formed by the points Q, A and B is an isosceles triangle

☛ Related Questions:

• Find the value of m if the points (5, 1), (–2, –3) and (8, 2m ) are collinear
• If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find . . . .
• Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9)

Question 2 Coordinated Geometry - Exercise 7.3

Next

Answer:

Point Q lies on the perpendicular bisector of AB, so the x-coordinate of the mid point of AB is the same as x-coordinate of point Q.

\text{By mid point formula,}\\ x-\text{coordinate}=\frac{\left(x_1 + x_2\right)}{2} = \frac{(-5+4)}{2}= -\frac{1}{2}\\ \text{Given that, }Q\text{ lies on }x\text{ axis, so }y=0.\\ P(x,y)= \left(-\frac{1}{2}, 0\right)

Therefore, it is an isosceles triangle.

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