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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

Respuesta :

Answer:

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 [tex]6x-11=y[/tex]

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

[tex]m_1 = \dfrac{y_B - y_A}{x_B - x_A}= \dfrac{1-(-1)}{2-14} =\dfrac{-1}{6}[/tex]

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

[tex]\dfrac{-1}{6}=\dfrac{1-y}{2-x}[/tex]

we solve the last equation for [tex]y[/tex] and so we get

[tex]y=\dfrac{-x}{6}+\dfrac{4}{3}[/tex]

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

[tex]y(0) = \frac{-0}{6}+\frac{4}{3}=\frac{4}{3}[/tex]

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

[tex]m_2 = -\dfrac{1}{m_1}=-\dfrac{1}{-\frac{1}{6}}=6[/tex]

Usinge the slope [tex]m_2[/tex] 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

[tex]6=\dfrac{y-1}{x-2}[/tex]

we solve the equation for y and we get the equation

[tex]y=6x-11[/tex]

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

[tex]13 = 6 x - 11 \quad \Rightarrow \quad x = \dfrac{13 + 11}{6}=\dfrac{24}{6}=4[/tex]

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

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).

(c) The y-coordinate of the point C is 13.

Calculation:

The equation of line passing through point [tex](x_{1},y_{1} )[/tex] is as follows:

[tex]y-y_{1} =m(x-x_{1})[/tex]                                                      ......(1)

Here, [tex]m[/tex] is the slope of the line.

The slope [tex]m[/tex] of the line passing through point [tex](x_{1} ,y_{1} )[/tex] and [tex](x_{2} ,y_{2})[/tex] is calculated as below:

[tex]m=\dfrac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex]                        ......(2)

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

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

[tex]m=\dfrac{-1-1}{14-2}\\m=\dfrac{-2}{12}\\m=-\dfrac{1}{6}[/tex]

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

[tex]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}[/tex]

To obtain the y-intercept substitute 0 for [tex]x[/tex] in above equation.

[tex]y(0)=-\dfrac{0}{6}+\dfrac{4}{3}\\y(0)=\dfrac{4}{3}[/tex]

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

[tex]m_{\text{BC}}=-\dfrac{1}{m_{\text{AB}}}[/tex]

Substitute [tex]-\dfrac{1}{6}[/tex] for [tex]m_{\text{AB}[/tex] in above equation.

[tex]m_{\text{BC}}=-\dfrac{1}{(-1/6)}\\m_{\text{BC}}=6[/tex]

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

Substitute 6 for [tex]m[/tex], 2 for [tex]x_{1}[/tex] and 1 for [tex]y_{1}[/tex] in equation (1).

[tex]y-1=6(x-2)\\y-1=6x-12\\y=6x-12+1\\y=6x-11[/tex]

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

[tex]13=6x-11\\6x=11+13\\6x=24\\x=4[/tex]

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

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)  

https://brainly.com/question/1332667  

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

https://brainly.com/question/2334270  

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.  

 

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