How do you write in point form (1,-5) perpendicular to - Xty =1

Answer 1

1. Find the Slope of the Given Line

First, rewrite the given equation in slope-intercept form (if necessary) to determine its slope. The given equation is:

−Xty=1-Xty = 1−Xty=1

This equation doesn't seem to be in a standard form for a line. Assuming the intention was to represent the line as:

y=−1Xy = -\frac{1}{X}y=−X1​

(though it's an unusual form), the slope of the line y=−1Xy = -\frac{1}{X}y=−X1​ (if XXX represents the x-axis) is −1X-\frac{1}{X}−X1​. For our purposes, we'll convert it to the more typical standard form of a line:

2. Convert to Standard Line Form

For a standard line equation in the form Ax+By=CAx + By = CAx+By=C, the general form of the line might be:

Ax+By=CA x + B y = CAx+By=C

For example, consider the equation:

x+y=1x + y = 1x+y=1

Here, the slope (mmm) is:

m=−ABm = -\frac{A}{B}m=−BA​

In the standard line equation, to find the slope of the line perpendicular to it, you need the negative reciprocal of the original line's slope.

3. Calculate the Perpendicular Slope

For a line with slope mmm, the slope of a perpendicular line is −1m-\frac{1}{m}−m1​.

4. Write the Equation of the Perpendicular Line

If the given equation was:

x+y=1x + y = 1x+y=1

The slope here is −1-1−1. The slope of the perpendicular line would be the negative reciprocal:

−1−1=1-\frac{1}{-1} = 1−−11​=1

So, the perpendicular line will have a slope of 111. If this perpendicular line passes through a point (x1,y1)(x_1, y_1)(x1​,y1​), its equation can be written as:

y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​)

Substitute the slope m=1m = 1m=1:

y−y1=1(x−x1)y - y_1 = 1(x - x_1)y−y1​=1(x−x1​)

Simplify to:

y=x−x1+y1y = x - x_1 + y_1y=x−x1​+y1​

5. Write in Point Form

If you are given a specific point through which the perpendicular line passes (let’s say (x1,y1)(x_1, y_1)(x1​,y1​)), plug these coordinates into the equation to get the specific line equation.