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Find the distance of the point $(7,1)$ from the line $3x+4y=4$ measured parallel to the line $3x-5y+2=0.$

I get that we get the perpendicular distance from the line very easily but what i dont understand is the phrase measured parallel to ...

I know this is a very elementary question...but i can solve it if i understand the language of the question.

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  • $\begingroup$ google.co.in/… $\endgroup$ Jan 17, 2018 at 7:28
  • $\begingroup$ A way you could solve this is to find the line that has the same gradient as $3x-5y=-2$ that passes through $(7,1)$. Then use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ $\endgroup$
    – Landuros
    Jan 17, 2018 at 7:35

2 Answers 2

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I think the problem could be understand in the following way (I will not provide the full computation since it seems you want to finish the task by yourself:) ).

The distance between a point $P$ and a given line $q$ is defined as the length of the smallest vector you can construct between $P$ and $q$. In other words, take arbitrary point $Q \in q$ and consider the vector $Q - P$. This vector has some length and the distance between $P$ and $q$ is the length of the smallest vector $Q - P$ (note: this vector is given uniquelly and is perpendicular to the line $q$). Now your problem is to find the distance parallel to some given line $q^{\prime}$. In other words, you want to measure the length of a vector that starts at $P$, ends at some point $Q \in q$ and the vector $Q - P$ is parallel to the line $q^{\prime}: 3x - 5y + 2 = 0$. From here you should be able to proceed without troubles.

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So by solving the two simultaneous equation

$3x+4y=4$

$3x-5y+2=0$

we get $y=1/3$ and $x=8/9$

So from here we can get by distance formula

$d^2$= $(x2−x1)^2+(y2−y1)^2$

we get $d$ = $6.147$.

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