Question

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

Answer 1

Hint:We need to know the concept & formula of finding distance between two points in the graph having different coordinates where, ordinate is the point in Y-axis and abscissa is the point in the X-axis. The distance of two points can be calculated using the formula. Here as the distance between between two points & coordinate of a point are given. As relation between abscissa & ordinate of other point is given so assume the point accordingly & form an equation by applying distance formula. Solve the equation to get coordinates of another point.

Complete step-by-step answer:
We know distance formula to find distance between two points,
$\operatorname{PQ} = \sqrt {{{\left( {{x_{_1}} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} $ , the two point being P$\left( {{x_1},{y_1}} \right)$and Q$\left( {{x_2},{y_2}} \right)$.
Let the coordinate of point A be $\left( {x,y} \right)$.
By the given information in the question , the ordinate of A is twice its abscissa.
$\therefore y = 2x$
Also, A is at a distance of $\sqrt {10} $ units from the point $\left( {4,3} \right)$. Hence, using the formula, we have
$\sqrt {10} = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {2x - 3} \right)}^2}} $
$\sqrt {10} = \sqrt {{x^2} - 8x + 16 + 4{x^2} - 12x + 9} $ [ As we know , \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]]
Now, squaring both sides,
$10 = {x^2} - 8x + 16 + 4{x^2} - 12x + 9$
$ \Rightarrow 5{x^2} - 20x + 15 = 0$ [Splitting middle term such that no.s should be multiple of 15 & 5]
$ \Rightarrow 5{x^2} - 15x - 5x + 15 = 0$ [ Taking (x-3) as common multiple]
${
   \Rightarrow 5x\left( {x - 3} \right) - 5\;\left( {x - 3} \right) = 0 \\
    \\
} $
$ \Rightarrow \left( {x - 3} \right)\;\left( {5x - 5} \right) = 0$
Now, $\left( {x - 3} \right) = 0$ $ \Rightarrow x = 3$
\[or,\left( {5x - 5} \right) = 0\] $ \Rightarrow x = \dfrac{5}{5} = 1$$ \Rightarrow x = \dfrac{5}{5} = 1$
 So , Either, $x = 3,\,\;i.e.y = 6$[As they given in question the ordinate of A is twice its abscissa].
OR, $x = 1,\;\;y = 2$
$\therefore $ The coordinate of point A can be $\left( {3,6} \right)\;\left( {1,2} \right)$.

So, the correct answer is “Option A”.

Note:Concept of Distance formula in two-dimensional coordinate geometry (having only two axis) should be clear to solve this problem. Ordinate and Abscissa should be put in the formula carefully and students should remember the distance between two points formula for solving these types of questions.