Question

Consider the line y=7/5x+4

Answer 1

The genral equation of line can be written as y=mx+c, where m is the slope of the line.

Comparing y=7/5x+4​ with y= mx+c, we get slope m=7/5.

The slope of two parallel lines are equal. Hence, the slope of a parallel line to y=7/5x+4​ is 7/5.

Now, the point slope form of a line which passes through a point (x1, y1) and habing slope m is,

`(y-y_1)/(x-x_1)=m`

Given (x1,y1)=(-7,-5). Therefore,

`\begin{gathered} (y-(-5))/(x-(-7))=(7)/(5) \\ (y+5)/(x+7)=(7)/(5) \\ 5y+25=7x+49 \\ 5y=7x+49-25 \\ 5y=7x+24 \\ y=(7)/(5)x+(24)/(5) \end{gathered}`

Therefore, the equation of line parallel to y=7/5x+4 and passing through point(-7,-5) is y=(7/5)x+(24/5).

(2)The slope of a line perpendicular to y=mx+c is -1/m.

So, slope of a line perpendicular to y=7/5x+4 is

`Slope,m_1=(-1)/(m)=(-1)/((7)/(5))=(-5)/(7)`

If (x1,y1)=(-7-5), then using point slope form,

`\begin{gathered} (y-y_1)/(x-x_2)=m_1 \\ (y-(-5))/(x-(-7))=(-5)/(7) \\ (y+5)/(x+7)=(-5)/(7) \\ 7y+35=-5x-35 \\ 7y=-5x-70 \\ y=(-5)/(7)x-10 \end{gathered}`

Therefore, the equation of line perpendicular to y=7/5x+4 and passing through point(-7,-5) is y=-(5/7)x-10.