Write the equation of a line that is perpendicular to y= 7/5 x +6 and that passes through the point 2,-6

Answer:The equation for line with point (2,-6) is given as : 5 x + 7 y = -32Step-by-step explanation:The given equation of line 1  is y= 7/5 x +6 Comparing it with the form  y = mx + C, we get m = 7/5.So, the slope of the equation 1 is ( 7/5).Now, let us assume the slope of line 2, which is perpendicular to line 1 is p.⇒ Slope of line 1 x Slope of line 2  = -1   (as lines are perpendicular)⇒ p x ( 7/5)  = -1⇒ p = - (5/7)Also, a point online 2 is given as ( 2,-6).By POINT SLOPE FORMULA:The line of the equation with point (x0,y0) and slope m is given as:y - y0 = m (x -x0)So, here the line equation (2) is given as[tex]y - (-6)  = \frac{-5}{7} (x -2)\\\implies 7(y +6) = -5(x-2)\\\implies 7y + 42  = - 5x + 10\\\implies 5x + 7y = -32[/tex]Hence, the equation for line 2 with point (2,-6) is given as : 5x + 7y = -32