If (-8, -4), (-2, 4) and (5, a) are collinear | vedclass

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(D) Three points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ are collinear if the slope of the line segment joining the first two points is equal to the

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$\frac{40}{3}$

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(D) Three points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ are collinear if the slope of the line segment joining the first two points is equal to the slope of the line segment joining the last two points.
Slope $m_1$ of the line segment joining $(-8, -4)$ and $(-2, 4)$ is given by $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-4)}{-2 - (-8)} = \frac{8}{6} = \frac{4}{3}$.
Slope $m_2$ of the line segment joining $(-2, 4)$ and $(5, a)$ is given by $m_2 = \frac{a - 4}{5 - (-2)} = \frac{a - 4}{7}$.
Since the points are collinear,$m_1 = m_2$.
Therefore,$\frac{4}{3} = \frac{a - 4}{7}$.
Multiplying both sides by $21$,we get $4 	imes 7 = 3(a - 4)$.
$28 = 3a - 12$.
$3a = 28 + 12 = 40$.
$a = \frac{40}{3}$.

If $(-8, -4), (-2, 4)$ and $(5, a)$ are collinear points,then find the value of $a$.

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