Question

length of perpendicular form (3,1) on line 4x+3y+20=0​

Answer 1

SOLUTION

TO DETERMINE

The length of perpendicular form (3,1) on line 4x+3y+20=0

CONCEPT TO BE IMPLEMENTED

The length of the perpendicular from

$\sf{(x_1, y_1)} \: on \: the \: line \: ax + by + c = 0 \: is$

$\displaystyle \sf{ = \bigg| \frac{ax_1 + by_1 + c}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg| }$

EVALUATION

Here the given equation of the line is

$\sf{4x + 3y + 20 = 0 \: \: \: ......(1)}$

Now the given point is ( 3, 1 )

Hence the required perpendicular distance

$\displaystyle \sf{ = \bigg | \: \frac{(4 \times 3) + (3 \times 1) + 20}{ \sqrt{ {4}^{2} + {3}^{2} } } \bigg| } \: unit$

$\displaystyle \sf{ = \bigg | \: \frac{12 + 3 + 20}{ \sqrt{ 16 + 9} } \bigg| }$

$\displaystyle \sf{ = \bigg | \: \frac{35}{ \sqrt{ 25} } \bigg| } \: \: unit$

$\displaystyle \sf{ = \bigg | \: \frac{35}{5 } \bigg| } \: \: unit$

$\displaystyle \sf{ = 7\: \: unit}$

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