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Find 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 $B (4, − 2)$ . Name the type of triangle formed by the points Q, A and B.

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Q (- \frac{1}{2}, 0)

, Isosceles Triangle

Q (- 1, 0)

, Right-angled Triangle

Q (\frac{1}{2}, 0)

, Equilateral Triangle

Q (1, 0)

, Scalene Triangle

answer is A.

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Detailed Solution

It is given that,
Point Q is on the

x

-axis and Q lies on the perpendicular bisector of the line AB.
The coordinates of two points which forms the line AB is

A − 5, − 2, B 4, − 2

.
Since,
Point Q is on the x-axis then the coordinate of point Q is

(x, 0)

.
Now point Q is equidistant from point A and B.
Then,

Q A = Q B

. … (1)
We know that, distance between points

(x, y) and (m . n)

is given by
Distance formula

= (m − x) 2 + (n − y) 2

.
Then,
The distance between point Q and A is,

Q A = (− 5 − x) 2 + (− 2 − 0) 2

⇒ Q A = x 2 + 10 x + 25 + 4 ⇒ Q A = x 2 + 10 x + 29

And
The distance between point Q and B is,

Q B = (4 − x) 2 + (− 2 − 0) 2

⇒ Q B = 16 − 8 x + x 2 + 4 ⇒ Q B = x 2 − 8 x + 20

Now,
From (1),

Q A = Q B ⇒ x 2 + 10 x + 29 = x 2 − 8 x + 20 ⇒ 10 x + 29 = 20 − 8 x ⇒ 18 x = − 9 ⇒ x = − 12

∴ The coordinates of point Q are − 12, 0

.
Since,

QA = QB

, hence we can say the two sides of

∆ QAB

is equal.
So,

∆ QAB

is an isosceles triangle.

∴ The coordinates of point Q are − 12, 0 and Q A B is an isosceles triangle .

Hence, option (1) is correct.

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