Question

the perpendicular distance of the point (3,-4) from the line 2x-5y+2=0 is​

Answer 1

Answer:

As you mention, points (-3,-4) to the line

3x-4y-1=0,

• let x=-3 and y=-4

• So comparing the line with equation

• Ax+By+C=0

• A=3, B=-4 and C=-1

Equation for distance between line and coordinate axes pts. Is

$d = \frac{ |ax + by + c| }{ \sqrt{(a {}^{2} } + b {}^{2}) }$

$d = \frac{ |3( - 3) - 4( - 4) - 1| }{ \sqrt{(9 + 16)} }$

d = 6/5

Hope it Helps !!

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Answer 2

$\dfrac{28}{\sqrt{29}}$ units

The perpendicular distance of the point $(3,-4)$ from the line $2x-5y+2=0$ is $\dfrac{28}{\sqrt{29}}$ units.

Concept:

The perpendicular distance of the point $(x_{1},y_{1})$ from the straight line $ax+by+c=0$ is given by

$d=\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}$

Step-by-step explanation:

Here the given straight line is

$2x-5y+2=0$

or, $(2)x+(-5)y+(2)=0$ ... ... (1)

So, the perpendicular distance of the point $(3,-4)$ from (1) no. straight line is

$\dfrac{|2(3)+(-5)(-4)+2|}{\sqrt{(2)^{2}+(-5)^{2}}}$ units

$=\dfrac{|6+20+2|}{\sqrt{4+25}}$ units

$=\dfrac{|28|}{\sqrt{29}}$ units

$=\dfrac{28}{\sqrt{29}}$ units

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