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2 years ago

Line AB and BC for a right angle at point B. If A(-3,4) and B(4,4) what is the equation of line BC?

Mathematics

2 answers:

Marta_Voda [28]2 years ago

8 0

Answer: The required equation of line BC is $x=4.$

Step-by-step explanation: As shown in the attached figure, lines AB and BC meet at right angle at the point B. The co-ordinates of point A and B are (-3, 4) and B(4, 4).

We are to find the equation of line BC.

We know that the slope of a line passing through the points (a, b) and (c, d) is given by

$m=\dfrac{d-b}{c-a}.$

So, the slope of the line AB will be

$m=\dfrac{4-4}{4-(-3)}\\\\\Rightarrow m=0.$

Therefore, the equation of the line AB is

$y-4=m(x-4)\\\\\Rightarrow y-4=0\\\\\Rightarrow y=4.$

Since y = constant is the equation of a line parallel to X-axis, so its perpendicular line will be parallel to Y-axis.

So, its equation will be of the form

x = constant.

Since the line BC is perpendicular to AB passing through the point (4, 4), so we must have

$x=4.$

Thus, the required equation of line BC is $x=4.$

andrey2020 [161]2 years ago

4 0

The correct answer is:

x=4.

Explanation:

The slope of line AB is given by the formula
 $m=\frac{y_2-y_1}{x_2-x_1}$ 

Using the coordinates of A and B, we have
 $m=\frac{4-4}{4--3}=\frac{0}{7}=0$ 

Perpendicular lines have slopes that are negative reciprocals. Since the slope of AB is 0/7, the reciprocal is 7/0, which is undefined.

Any line with an undefined slope is a vertical line. This will be a vertical line that passes through point B.

Vertical lines have the equation x=c, where c is a constant; since it runs through B, which has coordinates (4, 4), this means the equation is x=4.

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The plane we want to find has general equation

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$\vec n=a\,\vec\imath+b\,\vec\jmath+c\,\vec k$

$\vec n$ is perpendicular to both the normal vector of the other plane, which is $4\,\vec\imath+5\,\vec\jmath+6\,\vec k$ , as well as the tangent vector to the line $\vec r(t)$ , which is $\vec\imath+2\,\vec\jmath+9\,\vec k$ .

This means the dot product of $\vec n$ with either vector is 0, giving us

$\begin{cases}4a+5b+6c=0\\a+2b+9c=0\end{cases}$

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and we get

$(4a+5b)-4(a+2b)=-6-4(-9)\implies-3b=30\implies b=-10$

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$\boxed{11x-10y+z=-21}$

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2 years ago

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Reika [66]

The correct answer is:

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Explanation:

We will use the Rational Roots Theorem to solve this.

First we can divide both sides of the equation by 4 in order to simplify it:

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In order to solve this, we want the polynomial set equal to 0. To do this, subtract 105 from both sides:

$x^3-18x^2+80x-105=105-105 \\x^3-18x^2+80x-105=0$

The Rational Roots Theorem says that if p/q is a root of the polynomial, then p is a factor of the constant term and q is a factor of the leading coefficient. The constant term is -105. Drawing a factor tree, we find that the factors of this number are 1, -1, 3, -3, 5, -5, 7, -5, 15, -15, 21, -21. The leading coefficient is 1; its only factor is 1. This means p/q must be a whole number, and can be any of the factors of 105.

Using synthetic division, we try 1 in the box:

1 | 1 -18 80 -105

_______ 1__ -17___63__

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_______-1__ 19 _-99_

1 -19 99 -204

This is not a root; in fact, it shows us that none of the negatives will be a factor, as the absolute values increase as we complete the synthetic division.

Trying 3,

3 | 1 -18 80 -105

________3__-45___105_

1 -15 35 0

Since there is no remainder, 3 is a root, and is the answer we are looking for.

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