Answer

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**Hint:**As the points lie on the line segment that joins $(1,3)$ and $(2,7)$, therefore, their slope must have the same value and since their slopes are equal so from this we can easily find out the value of $a$.

**Formula used:**

Slope of the line $ = \dfrac{{y_2^{} - {y_1}}}{{{x_2} - {x_1}}}$

The section formula $\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}$

**Complete step by step answer:**

We have to find out the slope of the line joining the points $(1,3)$ and $(2,7)$ by applying the slope formula.

Here the vertices are

$\left( {{x_1},{y_1}} \right) = \left( {1,3} \right)$

$\left( {{x_2},{y_2}} \right) = \left( {2,7} \right)$

Therefore slope of the line $ = \dfrac{{y_2^{} - {y_1}}}{{{x_2} - {x_1}}}$$ = \dfrac{{7 - 3}}{{2 - 1}}$

On subtracting we get,

$ = 4$

Hence the slope of the line is $4$.

Since the slope will be same for both \[\left( {\dfrac{{5 + a}}{7},\dfrac{{6a + 3}}{7}} \right)\] and $(1,3)$.

Here the vertices are

$\left( {{x_1},{y_1}} \right) = \left( {1,3} \right)$

$\left( {{x_2},{y_2}} \right) = \left( {\dfrac{{5 + a}}{7},\dfrac{{6a + 3}}{7}} \right)$

Now using the formula for slope we get,

\[\Rightarrow \dfrac{{\dfrac{{6a + 3}}{7} - 3}}{{\dfrac{{5 + a}}{7} - 1}} = 4\]

By doing cross-multiplication we get

\[\Rightarrow \dfrac{{6a + 3}}{7} - 3 = 4\left( {\dfrac{{5 + a}}{7} - 1} \right)\]

By doing L.C.M we get

$\Rightarrow \dfrac{{6a + 3 - 21}}{7} = 4\left( {\dfrac{{5 + a - 7}}{7}} \right)$

Let us multiply the terms in RHS we get,

$\Rightarrow \dfrac{{6a + 3 - 21}}{7} = \dfrac{{20 + 4a - 28}}{7}$

Now $7$gets cancelled from both the sides and we get-

$\Rightarrow 6a + 3 - 21 = 20 + 4a - 28$

By solving it we get,

$\Rightarrow 2a = 10$

$\Rightarrow a = 5$

Thus the point \[\left( {\dfrac{{5 + a}}{7},\dfrac{{6a + 3}}{7}} \right)\]

We substitute the value of a becomes $\left( {\dfrac{{5 + 5}}{7},\dfrac{{6x5 + 3}}{7}} \right)$

On some simplification we get,

$ = \left( {\dfrac{{10}}{7},\dfrac{{33}}{7}} \right)$

Now we have to apply the section formula $\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$ to know the coordinates of the points which divides a line in two parts in the ratio of \[m:n\].

Therefore we can write that the coordinates are

Here the vertices are

$\left( {{x_1},{y_1}} \right) = \left( {1,3} \right)$

$\Rightarrow \left( {{x_2},{y_2}} \right) = \left( {2,7} \right)$

$\Rightarrow \left( {\dfrac{{2m + n}}{{m + n}},\dfrac{{7m + 3n}}{{m + n}}} \right) = \left( {\dfrac{{10}}{7},\dfrac{{33}}{7}} \right)$

So we can write the first term,

$\Rightarrow \dfrac{{2m + n}}{{m + n}} = \dfrac{{10}}{7}$

By doing cross-multiplication we get

$\Rightarrow 14m + 7n = 10m + 10n$

Now the \[m\] as LHS and \[n\] as RHS we get-

$\Rightarrow 4m = 3n$

So, $\dfrac{m}{n} = \dfrac{3}{4}$

**Thus, the required ratio is $3:4$ and the value of $a$ is $5$.**

**Note:**

While solving this question we need to keep in mind formulas related to a straight line in coordinate geometry and many of us make mistakes in this type of calculation.

The basic formula for finding out the slope of the line $ = \dfrac{{y_2^{} - {y_1}}}{{{x_2} - {x_1}}}$

And another important formula which you require in solving this question is$\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}$

Try to remember this formula otherwise the sum will be wrong.

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