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Svetllana [295]
3 years ago
11

1) find the equation of the line parallel to

Mathematics
1 answer:
In-s [12.5K]3 years ago
4 0

<u>Answer:</u>

1) The equation of the line parallel to  x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is  2y = 5x -12

<u>Solution:</u>

<u><em>1) find the equation of the line parallel to  x-5y=6 and through (4,-2).</em></u>

Given, line equation is x – 5y = 6  

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }

y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14

Hence, the line equation is 5y = x -14

<u><em> 2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)</em></u>

\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line \times slope of given line = -1

slope of required line = -1 \times \frac{5}{-2}=\frac{5}{2}

And it passes through (2, -1)

Now, using point slope form

\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}

Hence, the line equation is 2y = 5x -12

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