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# Find the ratio in which the points $\left( {\dfrac{{5 + a}}{7},\dfrac{{6a + 3}}{7}} \right)$ divides the join of $(1,3)$ and $(2,7)$. Also find the value of $a$.

Last updated date: 20th Jun 2024
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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.