What is the equation of the line that is perpendicular to the line 5x – 3y = 2 and passes through the point (-1,3)?

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03-11-2022
Mathematics

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What is the equation of the line that is perpendicular to the line 5x – 3y = 2 and passes through the point (-1,3)?

Sagot :

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SivanT355363 SivanT355363 · Answer 1 · 2022-11-03T04:21:21-04:00

Answer:

3x+5y=12.

Explanation:

Given the line: 5x-3y=2

First, we determine the slope by making y the subject of the equation.

[tex]\begin{gathered} 3y=5x-2 \\ y=\frac{5}{3}x-\frac{2}{3} \end{gathered}[/tex]

Comparing with the slope-intercept form: y=mx+b

• Slope = 5/3

Let the slope of the perpendicular line = n

By definition. two lines are perpendicular if the product of their slopes is -1.

Therefore:

[tex]\begin{gathered} \frac{5}{3}\times n=-1 \\ n=-\frac{3}{5} \end{gathered}[/tex]

Next, we use the point-slope form to find the perpendicular to the given line that is passing through (-1, 3).

[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-3=-\frac{3}{5}(x-(-1)) \\ y-3=-\frac{3}{5}(x+1)\text{ Multiply both sides by 5} \\ 5(y-3)=-3(x+1) \\ 5y-15=-3x-3 \\ 5y+3x=-3+15 \\ 3x+5y=12 \end{gathered}[/tex]

The required equation is 3x+5y=12.

Sagot :

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