Respuesta :
Question:
On a coordinate plane, a line goes through (negative 3, 2) and (0, 1). A point is at (3, 4). What is the equation of the line that is perpendicular to the given line and passes through the point (3, 4)? y = –One-thirdx + 5 y = –One-thirdx + 3 y = 3x + 2 y = 3x − 5
Answer:
Option D: [tex]y=3x-5[/tex] is the equation of the line
Explanation:
It is given that the line passes through the points [tex](-3,2)[/tex] and [tex](0,1)[/tex]
We shall find the equation of the line that passes through the points [tex](-3,2)[/tex] and [tex](0,1)[/tex]
Slope [tex]m=\frac{1-2}{0+3}[/tex]
[tex]m=\frac{-1}{3}[/tex]
Also, it is given that the line is perpendicular to the line having slope [tex]m=\frac{-1}{3}[/tex]
Since, we know that if two lines are perpendicular, then the product of the two slope is equal to -1.
Thus, we have,
[tex]m_1\cdot m_2=-1[/tex]
[tex]\frac{-1}{3} \cdot m_2=-1[/tex]
[tex]m_2=3[/tex]
Thus, the slope of the line is [tex]m=3[/tex] and passes through the point [tex](3,4)[/tex]
Now, we shall find the equation of the line using the formula,
[tex]y-y_1=m(x-x_1)[/tex]
Thus, we have,
[tex]y-4=3(x-3)[/tex]
[tex]y-4=3x-9[/tex]
[tex]y=3x-5[/tex]
Thus, the equation of the line is [tex]y=3x-5[/tex]
Hence, Option D is the correct answer.
