Question

The coordinates of the point $Q$ on the $x$-axis lies on the perpendicular bisector of the line segment joining the points $A(5,2)$ and $B(4,2)$. Name the type of triangle formed by the points $Q,A$ and $B$.

• A. Equilateral
• B. Isosceles
• C. Scalene
• D. None of these

Answer 1

Answer: (B)

Isosceles

The point $Q$ lies on the $x$-axis.

So, its co-ordinate will be of the form $(x,0)$

Let's take the points as $Q(x,y)=(x,0)$ which is equidistant from $A(5,2)$ and $B(4,2)$

To find the distance, we use distance formula

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$⟹AQ=\sqrt{(x-x_1{)}^2+(y-y_1{)}^2}=\sqrt{(x-5{)}^2+(0-2{)}^2}$

and $BQ=\sqrt{(x-x_2{)}^2+(y-y_2{)}^2}=\sqrt{(x-4{)}^2+(0-2{)}^2}$

But $AQ=BQ$

$\therefore \sqrt{{(x-5)}^2+{(0-2)}^2}=\sqrt{(x-4{)}^2+{(0-2)}^2}$

$⟹x=\frac{9}{2}=4.5$

$\therefore$ The co-ordinates of $Q$ is $(4.5,0)$

$\because AQ=BQ$ and $Q$ lies on the perpendicular bisector of $AB$

Therefore, the $\Delta ABQ$ is an isosceles triangle.