Mathematics, 17.07.2019 22:30 alyssaarnold230

Type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar.
ab and bc form a right angle at their point of intersection, b.
and the equation of bc
if the coordinates of a and b are (14,-1) and (2, 1), respectively, the y-intercept of ab is
isy= x+ o
if the y-coordinate of point c is 13, its x-coordinate is

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The y-intercept of AB is y=4/3, the x-coordinate of the poit C is 4, the equation of the line that goes through BC is 6x-11=y

Step-by-step explanation:

To find the y-intercept of AB we first have to find the equation of the line that goes through this segment.

We firt compute the slope of the segement which is given by

$m_1 = \dfrac{y_B - y_A}{x_B - x_A}= \dfrac{1-(-1)}{2-14} =\dfrac{-1}{6}$

Using this we can compute the equation of the line through AB using the point B=(2,1) and the formula

$\dfrac{-1}{6}=\dfrac{1-y}{2-x}$

we solve the last equation for and so we get

$y=\dfrac{-x}{6}+\dfrac{4}{3}$

The y-intercept of AB is the point in the line with x-coordinate equals to 0. Hence, the y-intercept is

$y(0) = \frac{-0}{6}+\frac{4}{3}=\frac{4}{3}$

Now, since the segments AB and BC are perpendicular, the slope of the line that goes through the segment BC is

$m_2 = -\dfrac{1}{m_1}=-\dfrac{1}{-\frac{1}{6}}=6$

Usinge the slope m_2 and knowing that the line goes through the point B=(2,1), to find the equation of the line through the segmen AC we use the following formula

$6=\dfrac{y-1}{x-2}$

we solve the equation for y and we get the equation

y=6x-11

Finally, to compute the x-coordinate of the point C we use the last equation and the fact that the y-coordinate is 13, it holds that

$13 = 6 x - 11 \quad \Rightarrow \quad x = \dfrac{13 + 11}{6}=\dfrac{24}{6}=4$

Answer from: Jlynbodmer

The y intercept of line AB is $\fbox{{4/3}}$ and x-coordinate of point C is $\fbox{4}$ .

Further explanation:

Given information:

(a) AB and BC forms a right angle at their point of intersection B.

(b) The coordinate of point A is (14,-1) and point B is (2,1).

Calculation:

The equation of line passing through point $(x_{1},y_{1} )$ is as follows:

$y-y_{1} =m(x-x_{1})$ ......(1)

Here, is the slope of the line.

The slope of the line passing through point $(x_{1} ,y_{1} )$ and $(x_{2} ,y_{2})$ is calculated as below:

$m=\dfrac{y_{2}-y_{1} }{x_{2} -x_{1} }$ ......(2)

Consider the point A(14,-1) as $\text{A}(x_{2} ,y_{2})$ and B(2,1) as $\text{B}(x_{1} ,y_{1})$ .

Now substitute 14 for $x_{2}$ , 2 for $x_{1}$ , -1 for $y_{2}$ and 1 for $y_{1}$ in equation (2) to obtain the slope of line AB.

$m=\dfrac{-1-1}{14-2}\\m=\dfrac{-2}{12}\\m=-\dfrac{1}{6}$

Substitute 2 for $x_{1}$ , 1 for $y_{1}$ and $-\dfrac{1}{6}$ for in equation (1) to obtain the equation of line AB.

$y-1=-\dfrac{1}{6}(x-2)\\y-1=-\dfrac{x}{6}+\dfrac{2}{6}\\y=-\dfrac{x}{6}+\dfrac{1}{3}+1\\y=-\dfrac{x}{6}+\dfrac{4}{3}$

To obtain the y-intercept substitute 0 for in above equation.

$y(0)=-\dfrac{0}{6}+\dfrac{4}{3}\\y(0)=\dfrac{4}{3}$

Since the line AB and BC are perpendicular to each other, the slope of line BC is obtained as follows:

$m_{\text{BC}}=-\dfrac{1}{m_{\text{AB}}}$

Substitute $-\dfrac{1}{6}$ for $m_{\text{AB}$ in above equation.

$m_{\text{BC}}=-\dfrac{1}{(-1/6)}\\m_{\text{BC}}=6$

The line BC passes through point B therefore, the equation of line BC is obtained as follows:

Substitute 6 for , 2 for $x_{1}$ and 1 for $y_{1}$ in equation (1).

$y-1=6(x-2)\\y-1=6x-12\\y=6x-12+1\\y=6x-11$

Since the y coordinate of point C is 13, to obtain the x-coordinate of point C substitute 13 for y in above equation.

$13=6x-11\\6x=11+13\\6x=24\\x=4$

Thus, the y intercept of line AB is $\fbox{{4/3}}$ and x-coordinate of point C is $\fbox{4}$ .

Learn more:

1. What is the y-intercept of the quadratic function f(x) = (x – 6)(x – 2)? (0,–6) (0,12) (–8,0) (2,0)

2. Which is the graph of f(x) = (x – 1)(x + 4)?

Answer details:

Grade: Middle school.

Subjects: Mathematics.

Chapter: Coordinate geometry.

Keywords: line, slope, equation, x-intercept, y-intercept, perpendicular, right angle, straight lines, x-value, y-value, point, intersection, y=mx+c, y-y1=m(x-x1), function, middle term split method, binomial, parabola, straight lines, quadratic, polynomial, function, expression, equation, coordinate geometry.