Question

The perpendicular distance of the line 3x + 4y - 5 =0 from the origin is​

Answer 1

Question :-

The perpendicular distance of the line 3x + 4y - 5 =0 from the origin is,

Answer :-

→ Perpendicular distance is 1 unit .

To Find :-

→ Perpendicular distance from ( 0,0)

Formula used :-

$\sf{d \: = \bigg| \frac{a \: x_1\: \: + b \:y_1 + c}{ \sqrt{ {a}^{2} + {b}^{2} } }\bigg| }$

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Step - by - step explanation :-

Given that ,

equation of line is 3x + 4y -5 = 0 ....(1)

Origin (0,0)

We know that ,

Standard equation of line

→ ax+ by + c = 9 .......(2)

Comparing (1) and (2)

We get,

→ a = 3 ,b = 4 and c = - 5

$\sf{points \: are \: ( \: 0 \: 0) }\\ \sf{ x_1 = 0 \: \: y_1 = 0}$

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Now substitute the value in given formula ,

$\rightarrow \sf{d \: = \bigg| \frac{3 \times 0 + 4 \times 0 - 5}{ \sqrt{ {3}^{2} + {4}^{2} } }\bigg| } \\ \\ \rightarrow \sf{d \: = \bigg| \frac{ - 5}{ \sqrt{25} }\bigg| } \\ \\ \rightarrow \sf{d \: = \bigg| \frac{ - 5}{5} \bigg| } \\ \\ \rightarrow \sf{ d \: = \bigg| - 1 \bigg| } \\ \\ \rightarrow \boxed{\sf{d \: = 1}}$

Perpendicular distance is 1 unit .

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