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**Hint:**Here in this question, we have to find a value or x term in the 2nd coordinate. Take the coordinates of two points. Call one point. Point A \[\left( {{x_1},{y_1}} \right)\] and make the other Point B \[\left( {{x_2},{y_2}} \right)\]. Know the distance formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]. And further simplify using the mathematical operation we can find the required value of a.

**Complete step-by-step solution:**

The distance between two points is the length of the interval joining the two points. If the two points lie on the same horizontal or same vertical line. In general the distance can be found by subtracting the coordinates that are not the same.

The distance between two points of the \[xy\] -plane can be found using the distance formula. An ordered pair \[\left( {x,{\text{ }}y} \right)\] represents co-ordinate of the point, where x-coordinate (or abscissa) is the distance of the point from the centre and y-coordinate (or ordinate) is the distance of the point from the centre.

Formula to find Distance Between Two Points in 2d plane. Consider two points, point A \[\left( {{x_1},{y_1}} \right)\] and point B \[\left( {{x_2},{y_2}} \right)\] on the given coordinate axis.

The distance between these points is given as: \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

Now consider the given two coordinates, point A\[\left( {3,14} \right)\] and point B\[\left( {a,5} \right)\] and distance \[d = 9\]units.

Substitute all this in formula then

\[ \Rightarrow \,\,9 = \sqrt {{{\left( {a - 3} \right)}^2} + {{\left( {5 - 14} \right)}^2}} \]

On squaring both sides we get

\[ \Rightarrow \,\,{9^2} = {\left( {a - 3} \right)^2} + {\left( {5 - 14} \right)^2}\]

Expand the RHS Using the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], then

\[ \Rightarrow \,\,{9^2} = {a^2} + {3^2} - 2 \cdot a \cdot 3 + {5^2} + {14^2} - 2 \cdot 5 \cdot 14\]

\[ \Rightarrow \,\,81 = {a^2} + 9 - 6a + 25 + 196 - 140\]

On simplification, we get

\[ \Rightarrow \,\,81 = {a^2} - 6a + 90\]

Take 81 to the RHS then

\[ \Rightarrow \,\,0 = {a^2} - 6a + 90 - 81\]

\[ \Rightarrow \,\,{a^2} - 6a + 9 = 0\]

You can finish from this point onwards by factorization. this is basically the same problem as factoring a quadratic equation.

Now, break the middle term as the summation of two numbers such that its product is equal to 9. Calculated above such two numbers are -3 and -3.

\[ \Rightarrow \,\,{a^2} - 3a - 3a + 9 = 0\]

Making pairs of terms in the above expression

\[ \Rightarrow \,\,\left( {{a^2} - 3a} \right) - \left( {3a - 9} \right) = 0\]

Take out greatest common divisor GCD from the both pairs, then

\[ \Rightarrow \,\,a\left( {a - 3} \right) - 3\left( {a - 3} \right) = 0\]

Take \[\left( {a - 3} \right)\] common

\[ \Rightarrow \,\,\,\left( {a - 3} \right)\left( {a - 3} \right) = 0\]

\[ \Rightarrow \,\,\,{\left( {a - 3} \right)^2} = 0\]

Take square root on both side, then

\[ \Rightarrow \,\,\,\left( {a - 3} \right) = 0\]

\[\therefore \,\,\,a = 3\] units

**Note:**The distance is a length between the two points. In the geometry we have a formula to determine the distance between the points. While determining the distance between the points we consider the both values of x and the value of y. Where x and y are the coordinates. the formula for distance is given by \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

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