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

Mathematics

2 answers:

umka2103 [35]3 years ago

6 0

Answer:

y = 1/3(x -2) +3

Step-by-step explanation:

The slope of the given line is the coefficient of x, -3. The slope of the perpendicular line will be the negative reciprocal of that: -1/-3 = 1/3. The line through a point (h, k) with slope m can be written in point-slope form as ...

y = m(x -h) +k

For m=1/3, (h, k) = (2,3), the equation of the line is ...

y = (1/3)(x -2) +3

Aleonysh [2.5K]3 years ago

3 0

Answer:

$\large\boxed{y=\dfrac{1}{3}x+2\dfrac{1}{3}}$

Step-by-step explanation:

$\text{Let}\ k:y=_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\============================\\\\\text{We have}\ y=-3x+1\to m_1=-3.\\\\\text{Therefore}\ m_2=-\dfrac{1}{-3}=\dfrac{1}{3}.\\\\\text{The equation of the searched line:}\ y=\dfrac{1}{3}x+b.\\\\\text{The line passes through }(2,\ 3).$

$\text{Put the coordinates of the point to the equation.}\ x=2,\ y=3:\\\\3=\dfrac{1}{3}(2)+b\\\\3=\dfrac{2}{3}+b\qquad\text{subtract}\ \dfrac{2}{3}\ \text{from both sides}\\\\b=2\dfrac{1}{3}$

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