Question

1. How do you write an equation of a line so that it is parallel or perpendicular to a given point and a given line?

Answer 1

@Nurali

Answer 2

y=mx+b parallel line equation

Answer 3

Since we are not given any, let's take general things:
Let's say we hace a line with equation y=(m1)x+b and a point T(k,n) wich is a point the parallel line passes through.
By property of parallel lines, the slope of the first slope must be the same than the given.
\[m_1 = m_2\]
Now we know that the second equation goes through T(k.n).
Then applying the equation:
\[(y-y_1 ) = m(x-x_1)\]
We know that m1 is equal to m2, and T(k.n) is a point, then:
\[(y-n)=m_2 (x-k)\]
Now, just simplify:
\[y=m_2 x - m_2(x_1)-n\]
must be the equation of the line parallel to the generalized line that was given, and the point also given.
So basically I just deduced the slope with the property and transformed it to an equation using the given point and the equation that relates a slope and a point.

Answer 4

You just identify the slope of the 1st line. Use that slope for parallel. Use the negative of that slope for perpendicular. Then, knowing slope and line, use point-slope form to build your second line.

If I have the equation: \[2x=3y-9\]

First I will solve it for y, to find m, my slope:
\[3y = 2x + 9\\y = \frac{2x+9}{3}\\y = \frac{2}{3}x + 3\]

So you know your slope, m, is 2/3.

Now, you have another point that this 2nd line goes through. Let's say it's \[(1,1)\]

Using point-slope form:

\[(y-y_1) = m(x-x_1)\]

\[y - 1 = \frac{2}{3}(x-1)\]

Simplified:

\[y = \frac{2}{3}x - \frac{2}{3} + 1\]
\[y = \frac{2}{3}x + \frac{1}{3}\]

This is the line that is parallel to the 1st, going through point (1,1).