Question

a straight line passes through the points (5,0) & (0,3) the length of the perpendicular from the points (4,4) on the line is

Answer 1

$eq. \: of \: line \\ (y - 0) = \frac{3 - 0}{0 - 5} (x - 5) \\ - 5y = 3x - 15 \\ 3x + 5y - 15 = 0 \\ leangth \: of \: perpendicular \: from \: (4 \: \: \: 4) \\ = | \frac{3 \times 4 + 5 \times 4 - 15}{ \sqrt{ {3}^{2} + {5}^{2} } } | \\ = | \frac{12 + 20 - 15}{ \sqrt{9 + 25} } | \\ = | \frac{32 - 15}{ \sqrt{34} } | \\ = \frac{17}{ \sqrt{34} } \\ = \frac{17}{ \sqrt{34} } \times \frac{ \sqrt{34} }{ \sqrt{34} } \\ = \frac{ \sqrt{34} }{2}$

Answer 2

Answer:

The length of the perpendicular from the points (4,4) on the line is 2.915 unit.

Step-by-step explanation:

We know that, the equation of straight line passes through the points $(x_{1} ,y_{1} )$ and $(x_{2} , y_{2} )$ :

$Y-y_{1} = m (X-x_{1} )$ ---------------------(1)

Where,   $m = \frac{y_{2}- y_{1} }{x_{2} - x_{1} }$ = slope of the line

According to the question, the straight line passes through the points (5,0) & (0,3). Here, $x_{1} = 5, y_{1} = 0, x_{2} = 0, y_{2} = 3.$

So, the slope of the line,$m_{1} = \frac{3-0}{0-5} = \frac{-3}{5}$

From equation (1),

$Y-0 = \frac{-3}{5} (X - 5)$

⇒$5Y = -3X + 15$

⇒$3X + 5Y = 15$ ----------------------(2)

We know that, the condition for two line which are perpendicular to each other: $m_{1}\times m_{2} = -1$

$(\frac{-3}{5} )\times m_{2} = -1$

⇒$m_{2} = \frac{5}{3}$

So, the equation of perpendicular line passes through (4,4) is,

$Y -4 = \frac{5}{3} (X - 4)$

⇒$3Y - 12 = 5X - 20$

⇒$5X - 3Y = 8$--------------------(3)

After solving the equation(2) and (3), we get

$X = \frac{5}{2}$ & $Y = \frac{3}{2}$

So, $(\frac{5}{2} , \frac{3}{2} )$ is the intersection point of both perpendicular lines.

We know that, the distance between two points is

$d = \sqrt{(x_{2} - x_{1} ) ^{2} + (y_{2}- y_{1} ) ^{2} }$

⇒$d = \sqrt{(\frac{5}{2}-4) ^{2}+(\frac{3}{2}-4) ^{2} }$

⇒$d = \sqrt{\frac{9}{4} + \frac{25}{4} }$

⇒$d = \sqrt{\frac{34}{4} } = 2.915$

So,  the length of the perpendicular from the points (4,4) on the line is 2.915 unit.