Lines p, q, and r all pass through point (-3,4). Line p has slope 4 and is perpendicular to line q. Line r passes through Quadrants 1 and Quadrants 2 only. Write an equation for each line. Then graph the three lines on the same coordinate plane.

Respuesta :

Answer:

Part 1) The equation of line p is [tex]y=4x+16[/tex]

Part 2) The equation of line q is [tex]y=-\frac{1}{4}x+\frac{13}{4}[/tex]

Part 3) The equation of line r is [tex]y=4[/tex]

The graph in the attached figure

Step-by-step explanation:

step 1

Find the equation of the line p

we have

[tex]m=4\\point\ (-3,4)[/tex]

Find the equation in point slope form

[tex]y-y1=m(x-x1)[/tex]

substitute

[tex]y-4=4(x+3)[/tex] ----> equation in point slope form

Convert to slope intercept form

[tex]y-4=4x+12[/tex]

[tex]y=4x+12+4[/tex]

[tex]y=4x+16[/tex]

step 2

Find the equation of line q

we know that

If two lines are perpendicular, then their slope are opposite reciprocal (the product of their slopes is equal to -1)

we have that

Line q is perpendicular to line q

so

[tex]m_q*m_p=-1[/tex]

we have

[tex]m_p=4[/tex]

therefore

[tex]m_q=-\frac{1}{4}[/tex]

we have

[tex]m=-\frac{1}{4}\\point\ (-3,4)[/tex]

Find the equation in point slope form

[tex]y-y1=m(x-x1)[/tex]

substitute

[tex]y-4=-\frac{1}{4}(x+3)[/tex] ----> equation in point slope form

Convert to slope intercept form

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

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

[tex]y=-\frac{1}{4}x+\frac{13}{4}[/tex]

step 3

Find the equation of line r

we know that

Line r passes through Quadrants 1 and Quadrants 2 only and passes through the point (-3,4)

therefore

The equation of Line r must be equal to the y-coordinate of the given point

[tex]y=4[/tex] ----> is a horizontal line

step 4

Graph the three lines

using a graphing tool

see the attached figure

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