Mathematics: Hi again I need help on 13&14 15&16

Hi again I need help on 13&14 15&16 - id131415165626267, GabbyGood1072

elainestorm9633

13.01.2023

ANSWER:

The line equation of required line is 5x – 6y – 35 = 0.

SOLUTION:

Given, line equation is 5x = 6y + 7 and the point is p (1, -5).

We have to find the line equation of a line that is parallel to given line and passing through point p.

First, let us find slope of given line.

5x = 6y + 7

5x – 6y – 7 = 0

$\text { Slope of a line }=\frac{-x \text { coefficient }}{y \text { coefficient }}$

$=\frac{-5}{-6}=\frac{5}{6}$

We know that, parallel lines will have same slope, so slope of required line is $\frac{5}{6}$

Now, we have slope and a point through it.

So, let us find the point slope form of the line i.e $y-y_{1}=m\left(x-x_{1}\right)$

$\text { Here, } \mathrm{y}_{1}=-5, \mathrm{x}_{1}=1 \text { and } \mathrm{m}=\frac{5}{6}$

Line equation → $y-(-5)=\frac{5}{6}(x-1)$

$y+5=\frac{5}{6}(x-1)$

6y + 30 = 5x – 5

5x – 6y – 5 – 30 = 0

5x – 6y – 35 = 0

Hence, the line equation of required line is 5x – 6y – 35 = 0.

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