Suppose m is in the line given by the equation 6x-3y=7, and suppose n is the line perpendicular to m and passing Nd through the point (6,2). If k is the line of slope 5 and y-intercept 1, what is the x-coordinate of the intersection of n and k? Express your answer in a common fraction.

Answer: The x co-ordinate o intersection of line k and n is [tex]\frac{8}{11}[/tex] Step-by-step explanation:Given as :The equation of line m is 6 x - 3 y = 7So, in the standard form , line equation isy = a x + c  , where a is the slopeSo,  6 x - 3 y = 7 can be written as 3 y = 6 x - 7or,  y = 2 x - [tex]\frac{7}{3}[/tex]       ........1So, slope of this line = a = 2Now, The line n is perpendicular to line m and passing through line ( 6 , 2 )So, Slope of line n = bFor , perpendicular lines , products of slope = - 1Or, a × b = -1∴  b = - [tex]\frac{1}{a}[/tex]I.e b = - [tex]\frac{1}{2}[/tex]So,equation of line n with slope b and passing through line ( 6 , 2 ) is y - [tex]y_1[/tex] = b ( x - [tex]x_1[/tex] )or, y - 2 =  - [tex]\frac{1}{2}[/tex] ( x - 6 )or, 2 × ( y - 2 ) =  - 1 ( x - 6 )or, 2 y - 4 = - x + 6or, x + 2 y -10 = 0          ........2Again, equation of line k with slope 5 and y intercept = 1For y intercept , x coordinate = 0y = c x + cor, 1 = c× ( 0 ) + cOr, c = 1Or, equation of line k isy = 5 x + 1           ..........3Now intersection of line k and n isput the value of y from eq 3 into eq 2I.e x + 2 × (  5 x + 1  )-10 = 0  Or, x + 10 x + 2 - 10 = 0or, 11 x - 8 = 0or 11 x = 8∴  x = [tex]\frac{8}{11}[/tex] Hence The x co-ordinate o intersection of line k and n is [tex]\frac{8}{11}[/tex]  Answer