Question

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 1

Answer:

The x co-ordinate o intersection of line k and n is $\frac{8}{11}$

Step-by-step explanation:

Given as :

The equation of line m is 6 x - 3 y = 7

So, in the standard form , line equation is

y = a x + c  , where a is the slope

So,  6 x - 3 y = 7 can be written as

3 y = 6 x - 7

or,  y = 2 x - $\frac{7}{3}$       ........1

So, slope of this line = a = 2

Now, The line n is perpendicular to line m and passing through line ( 6 , 2 )

So, Slope of line n = b

For , perpendicular lines , products of slope = - 1

Or, a × b = -1

∴  b = - $\frac{1}{a}$

I.e b = - $\frac{1}{2}$

So,equation of line n with slope b and passing through line ( 6 , 2 ) is

y - $y_1$ = b ( x - $x_1$ )

or, y - 2 =  - $\frac{1}{2}$ ( x - 6 )

or, 2 × ( y - 2 ) =  - 1 ( x - 6 )

or, 2 y - 4 = - x + 6

or, x + 2 y -10 = 0          ........2

Again, equation of line k with slope 5 and y intercept = 1

For y intercept , x coordinate = 0

y = c x + c

or, 1 = c× ( 0 ) + c

Or, c = 1

Or, equation of line k is

y = 5 x + 1           ..........3

Now intersection of line k and n is

put the value of y from eq 3 into eq 2

I.e x + 2 × (  5 x + 1  )-10 = 0  

Or, x + 10 x + 2 - 10 = 0

or, 11 x - 8 = 0

or 11 x = 8

∴  x = $\frac{8}{11}$

Hence The x co-ordinate o intersection of line k and n is $\frac{8}{11}$  Answer