What is the equation of the line perpendicular to $y = - \frac{25}{3} x$ $y=-25/3x$ that passes through $(- 1, - 6)$ $(-1,-6)$ ?

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Answer 1 · 2018-05-08T11:53:53

Equation of the line is $3 x - 25 y = 147$ $3 x - 25 y =147$

The slope of the line $y = - \frac{25}{3} x [y = m x + c]$ $y = - 25/3 x [y= m x+c ]$

is $m_{1} = - \frac{25}{3}$ $m_1= -25/3$ . The product of slopes of the perpendicular lines

is $m_1*m_2=-1 :. m_2 = (-1)/(-25/3)= 3/25$

The slope of the line passing through $(-1,-6)$ is $3/25$

The equation of line passing through $(x_1,y_1)$ having slope of

$m$ is $y-y_1=m(x-x_1)$ .

The equation of line passing through $(-1, -6)$ having slope of

$3/25$ is $y+6=3/25(x+1) or 25 y +150 = 3 x+3$ . or

$3 x - 25 y =147$

The equation of line is $3 x - 25 y =147$ [Ans]

What is the equation of the line perpendicular to y=-25/3x y=−253xy=-25/3x that passes through (-1,-6) (−1,−6) (-1,-6) ?