What is the equation of the line perpendicular to $y = \frac{2}{15} x$ $y=2/15x$ that passes through $(- 4, 4)$ $(-4,4)$ ?

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Answer 1 · 2018-05-21T11:00:23

Equation of the line is $y = - \frac{15}{2} x - 26$ $y = -15/2 x -26$

Slope of the line, $y = \frac{2}{15} x; [y = m x + c]$ $y=2 /15 x ; [y=m x+c]$

is $m_{1} = \frac{2}{15}$ $m_1= 2/15$ [Compared with slope-intercept form of equation]

The product of slopes of the pependicular lines is $m_{1} \cdot m_{2} = - 1$ $m_1*m_2=-1$

$:.m_2= -15/2$ . The equation of line passing through

$(x_1,y_1)$ having slope of $m_2$ is $y-y_1=m_2(x-x_1)$ .

The equation of line passing through $(-4,4)$ having slope of

$-15/2$ is $y-4=-15/2(x+4) or y = -15/2 x +4-30$ . or

$y = -15/2 x -26$ .

Equation of the line is $y = -15/2 x -26$ [Ans]

What is the equation of the line perpendicular to y=2/15x y=215xy=2/15x that passes through (-4,4) (−4,4) (-4,4) ?