225k views
3 votes
a graphic designer is creating a logo for a client. Lines DB and AC are perpendicular.The equation of DB is 1/2x+2y=12 what is the equation of AC

a graphic designer is creating a logo for a client. Lines DB and AC are perpendicular-example-1
User TenG
by
8.5k points

2 Answers

4 votes

Answer: -4x + y= -28

Explanation:

12⁢x+2⁢y=12y=-14⁢x+6The slope of line DB is -14. The slopes of perpendicular lines are opposite reciprocals of each other, so line AC will have a slope of 4. Substitute the slope of 4 and the point (8,4) into the slope-intercept formula to find the equation of line AC.4=4⁢(8)+b4=32+b-28=bSo, line A⁢C has a y-intercept, b, of -28 and a slope of 4. Convert the equation to standard form.

User Ian Pugsley
by
8.6k points
2 votes

y=4x-28

Explanation

when two lines are perpendicular the product o their slopes equals, -1


\begin{gathered} if \\ line_1\perp\text{ line}_2 \\ then \\ slope_1*slope_2=-1 \end{gathered}

so

Step 1

find the slope of the line AC

a)the function of the line DB is given in the form


\begin{gathered} y=mx+b \\ where \\ m\text{ is the slope} \\ b\text{ is the y-intercept} \end{gathered}

so


(1)/(2)x+2y=12

b)isolate y to get the slope-intercept form, and check the m value


\begin{gathered} (1)/(2)x+2y=12 \\ subtract\text{ }(1)/(2)x\text{ in both sides} \\ (1)/(2)x+2y-(1)/(2)x=12-(1)/(2)x \\ 2y=-(1)/(2)x+12 \\ divide\text{ both sides by 2} \\ (2y)/(2)=(-(1)/(2)x+12)/(2) \\ y=-(1)/(4)x+6\Rightarrow line\text{ 1} \\ so \\ slope_1=-(1)/(4) \end{gathered}

c) now, as the lines AC and DB are perpendicular replace to find slope 2


\begin{gathered} slope_1*slope_2=-1 \\ -(1)/(4)*slope_2=-1 \\ Multiply\text{ both sides by -4} \\ -(1)/(4)slope_2*-4=-1*-4 \\ slope_2=4 \end{gathered}

therefore

the slope of the line AC is 4

Step 2

now, we can use the slope-point formula to find the equaiton of the line,it says


\begin{gathered} y-y_1=m(x-x_1) \\ where\text{ m is the slope} \\ (x_1,y_1)\text{ is a point from the line} \end{gathered}

so

a)let


\begin{gathered} slope=4 \\ point=(8,4)\text{ this is a common point for both lines} \end{gathered}

b) finally, replace and solve for y


\begin{gathered} y-y_(1)=m(x-x_(1)) \\ y-4=4(x-8) \\ y-4=4x-32 \\ add\text{ 4 in both sides} \\ y-4+4=4x-32+4 \\ y=4x-28 \end{gathered}

so, the equation of the line AC ia

y=4x-28

I hope this helps you

User Jayendra Parmar
by
7.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories