If the equation of JK is y = mx + c, then y = mx + c' is the equation of the straight line parallel to above and $y = - \frac{1}{m}x + b$ 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

⇒ $n = - \frac{1}{m}$

Therefore, the equation of a straight line which is perpendicular to equation (1) will be $y = - \frac{1}{m}x + b$ where b is any real constant. (Answer)

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