Use the graph to answer the question. 5 3 1 5 4 3 2 1 EN 4 5 2 N -3 -4 TVhich line is perpendicular to the line shown on the graph? O À y=-x+2 Oc y=x+1 OD. y = 2 x + 1 +1 B. y=-37-4

Use the graph to answer the question 5 3 1 5 4 3 2 1 EN 4 5 2 N 3 4 TVhich line is perpendicular to the line shown on the graph O À yx2 Oc yx1 OD y 2 x 1 1 B y3 class=

Respuesta :

[tex]A)y=-\frac{3}{2}x+2[/tex]

Explanation

Step 1

find the slope of the given line

you can find the slope using:

[tex]\begin{gathered} \text{slope}=m=\frac{change\text{ in y }}{\text{change in x}}=\frac{y_2-y_1}{x_2-x_1} \\ \text{where} \\ P1(x_1,y_1) \\ P2(x_2,y_2) \end{gathered}[/tex]

P1 and P2 are 2 known points of the line

then,let

P1(0,-1)

P2(3,1)

replace

[tex]\begin{gathered} m_1=\frac{change\text{ in y }}{\text{change in x}}=\frac{y_2-y_1}{x_2-x_1} \\ m_1=\frac{1-(-1)}{3-0}=\frac{2}{3} \end{gathered}[/tex]

Step 2

Now, 2 lines are perpendicular if the product of their slopes equals - 1

[tex]\begin{gathered} m_1\cdot m_2=-1 \\ \text{Let} \\ m_1=\frac{2}{3} \\ \text{replace} \\ \frac{2}{3}\cdot m_2=-1 \\ \text{Multiply both sides by 3/2} \\ \frac{2}{3}\cdot m_2\cdot\frac{3}{2}=-1\cdot\frac{3}{2} \\ m_2=-\frac{3}{2} \end{gathered}[/tex]

Hence, the line we are looking for has a slope of -3/2

[tex]\begin{gathered} y=mx+b \\ \text{where} \\ m\text{ is the slope} \end{gathered}[/tex]

the number with the variable is the slope, so find in the optiions the answer with :

[tex]-\frac{3}{2}x[/tex]

so, the answer is

[tex]A)y=-\frac{3}{2}x+2[/tex]

ACCESS MORE
EDU ACCESS
Universidad de Mexico