What are the coordinates of the vertex of the function y-11=-(x-5)^2?
1 answer:
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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 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
⇒
Therefore, the equation of a straight line which is perpendicular to equation (1) will be where b is any real constant. (Answer)