Question

What is the equation of the line parallel to 3x + 5y = 11 that passes through the point (15, 4)?3x-29= 3x - 295y-y-x-21y--x+533y=-5* +13Mart this and returnSavond uit

Answer 1

D

Explanation:

The slope of the the straight line equation 3x+5y=11 is (−35)Hence the slope of the line parallel is same to the given line i.e,(−35) We can write the equation in the following form slope intercept form i.e, y=mx+cHere c=−6 and m=−35The answer is y=−35x−6⇒5y=−3x−30⇒3x+5y=−30Hope it helps...Thanks you...

Answer 2

Recall that two equations in standard form represent parallel lines if they are as follows:

`\begin{gathered} Ax+By=C_1, \\ Ax+By=C_2, \end{gathered}`

Where A>0, and all the coefficients are integers.

Therefore the equation of a parallel line to the given line is as follows:

`3x+5y=k`

Since the parallel line passes through (15,4) then:

`3*15+5*4=k.`

Simplifying the above result we get:

`\begin{gathered} 45+20=k, \\ k=65. \end{gathered}`

Therefore:

`3x+5y=65.`

Solving the above equation for y we get:

`\begin{gathered} 3x+5y-3x=65-3x, \\ 5y=-3x+65, \\ (5y)/(5)=-(3x)/(5)+(65)/(5), \\ y=-(3)/(5)x+13. \end{gathered}`

Answer: Last option.