Given A(-1,4), B(2,-5), and C(3, 4), which coordinate pair will make AB perpendicular to CD ?

Answer:

The coordinate pair of the point D that will make AB perpendicular to CD is the D(6, 3)

Step-by-step explanation:

The slope of the given line AB can be presented as follows;

[tex]Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]m =\dfrac{-5-4}{2-(-1)} = \dfrac{-9}{3} =-3[/tex]

The slope of a perpendicular line  to a given line is equal to the negative reciprocal of the slope of the given line

Therefore, the slope of the perpendicular to a line that has a slope of m is -1/m

The slope of the perpendicular to the line AB is therefore [tex]-\dfrac{1}{3}[/tex]

From which we get the equation of the perpendicular to the line AB given as follows;

y - 4 = -1/3 × (x - 3)

y - 4 = -x/3 + 1

y = -x/3 + 1 + 4

y = -x/3 + 5

When x = 6, y = -6/3 + 5 = 3

Therefore, the coordinate pair of the point D that will make AB perpendicular to CD is the D(6, 3).

Using the slope concept, it is found that the coordinate pair is:

[tex]D\left(x, \frac{x}{3}+9\right)[/tex]

The slope of a coordinate pair [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by the change in y divided by the change in x, that is:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

If two segments are perpendicular, the multiplication of their slopes is -1.

For segment AB, the slope is:

[tex]m = \frac{-5 - 4}{2 - (-1)} = -\frac{9}{3} = -3[/tex]

At segment CD, we want that:

[tex]-3m = -1[/tex]

[tex]m = \frac{1}{3}[/tex]

Considering D(x,y), we have that:

[tex]\frac{y - 4}{x - 3} = \frac{1}{3}[/tex]

Writing y as a function of x:

[tex]x - 3 = 3(y - 4)[/tex]

[tex]3y - 12 = x - 3[/tex]

[tex]3y = x + 9[/tex]

[tex]y = \frac{x}{3} + 9[/tex]

Thus, the coordinate pair is:

[tex]D\left(x, \frac{x}{3}+9\right)[/tex]

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