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(04.02 mc) write an equation of a line perpendicular to line ab in slop…

Question

(04.02 mc)
write an equation of a line perpendicular to line ab in slope - intercept form that passes through the point (7, 6).
options:
○ ( y = 0.5x + 25 )
○ ( y = 0.5x - 25 )
○ ( y = - 2x + 20 )
○ ( y = 2x - 20 )

Explanation:

Step1: Find slope of line AB

Identify points A$(0, -1)$ and B$(5, 1)$. Use slope formula:
$$m_{AB} = \frac{1 - (-1)}{5 - 0} = \frac{2}{5} = 0.4$$

Step2: Find perpendicular slope

Perpendicular slope is negative reciprocal:
$$m_{\perp} = -\frac{1}{m_{AB}} = -\frac{5}{2} = -2.5$$

Step3: Solve for y-intercept b

Use point $(7,6)$ and $y=mx+b$:
$$6 = -2.5(7) + b$$
$$6 = -17.5 + b$$
$$b = 6 + 17.5 = 23.5$$
*Correction: Recheck perpendicular slope rule: If original slope is $\frac{2}{5}$, perpendicular slope is $-\frac{5}{2}=-2.5$. Wait, match options: recheck line AB slope: A$(0,-1)$, B$(5,1)$: $\frac{1 - (-1)}{5-0}=\frac{2}{5}$. Perpendicular slope is $-\frac{5}{2}$? No, wait: perpendicular slope product is -1. $\frac{2}{5} \times m = -1 \implies m=-\frac{5}{2}=-2.5$. Wait no, options have $y=-2x+20$. Wait recheck points: A is $(0,-1)$? No, looking at graph: A is $(1,0)$, B is $(5,2)$. Oh! Correct points: A$(1,0)$, B$(5,2)$.

Step1 (corrected): Slope of AB

$$m_{AB} = \frac{2-0}{5-1} = \frac{2}{4} = 0.5$$

Step2 (corrected): Perpendicular slope

$$m_{\perp} = -\frac{1}{0.5} = -2$$

Step3 (corrected): Find b with (7,6)

$$6 = -2(7) + b$$
$$6 = -14 + b$$
$$b = 6 + 14 = 20$$

Step4: Form equation

$$y = -2x + 20$$

Answer:

$\boldsymbol{y = -2x + 20}$ (Option: y = -2x + 20)