Question

A point A is at a distance of $\sqrt{10}$ units from the point (4,3). Find the coordinates of point A, if its ordinate is twice its abscissa.
A. (3,6), (1,2)
B. (2,4), (2,3)
C. (6,12), (9,18)
D. none of these

Answer 1

Hint: The given question involves two different concepts of mathematics. First, we will have to use distance formulas between the two points and then we will have to use the methods to solve a quadratic equation.

Formula used:
Distance formula:
$d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_1)}^2}$
d is the distance two points $(x_1,y_1)$ and $(x_2,y_2)$

Complete step by step solution:
It is given that a point A is at a distance of $\sqrt{10}$ units from the point (4,3) on a Cartesian plane and we have to find the coordinates of these points A. It is also given that the ordinate of the point A is twice its abscissa.To find the coordinates of point A we shall use the formula for distance between two points (i.e. the distance formula).Let the coordinates of point A be (x , y) and let the coordinate of point B be (4 , 3). On applying the distance formula for the points A and B we get that,
$\sqrt{10}=\sqrt{{(x-4)}^2+{(y-3)}^2}$

Let us now square the both sides of the above equation.
With this we get that $10={(x-4)}^2+{(y-3)}^2$ … (i)
But we know that $y=2x$
Substitute the value of y in equation (i).
$\Rightarrow 10={(x-4)}^2+{(2x-3)}^2$.
Now, we shall open up the brackets and simplify the equation.
$\Rightarrow 10=x^2-8x+16+4x^2-12x+9$
$\Rightarrow 5x^2-20x+15=0$
On dividing the equation by 5 we get,
$x^2-4x+3=0$

Let us now solve for the equation. This equation can be written as $x^2-x-3x+3=0$.
$$\begin{gathered} x(x-1)-3(x-1)=0 \\ \Rightarrow (x-1)(x-3)=0 \end{gathered}$$
This means that either $(x-1)=0$ or $(x-3)=0$.This further means that $x=1$ or $x=3$.
If $x=1$, then $y=2x=2(1)=2$. Therefore, the coordinates of A are (1,2).If $x=3$, then $y=2x=2(3)=6$. Therefore, the coordinates of A are (3,6).

Hence, the correct option is A.

Note: From this question we get to know that there exist two different points on a cartesian plane that are equal distances from a fixed point. We solved the quadratic equation in x by factoring the expression. There are two other methods for solving a quadratic equation in one variable and that are completing the square method and quadratic formula.