Answer:
If the equation of JK is y = mx + c, then y = mx + c' is the equation of the straight line parallel to above and [tex]y = - \frac{1}{m}x + b[/tex] will be perpendicular to that.
Step-by-step explanation:
Let us assume that the line JK has the equation in slope-intercept form as
y = mx + c ............. (1)
Therefore, the equation has slope = m
Now, any straight line having an equation with slope m will be parallel to line JK.
So, the equation of the straight line which is parallel to equation (1) will be
y = mx + c', where, c' is any real constant.
Now, let us assume that another straight line having equation
y = nx + a is perpendicular to the line JK i.e. equation (1).
Now, we know if two lines are perpendicular to each other then the product of their slopes will be - 1.
So, mn = - 1
⇒ [tex]n = - \frac{1}{m}[/tex]
Therefore, the equation of a straight line which is perpendicular to equation (1) will be [tex]y = - \frac{1}{m}x + b[/tex] where b is any real constant. (Answer)
Answer:
You havent showed us a picture of the graph so how are we going to help you
Step-by-step explanation:
