Question

question 3 of 5
\overleftrightarrow{ab} and \overleftrightarrow{bc} form a right angle at point b. if a = (-3, -1) and b = (4, 4), what is the equation of \overleftrightarrow{bc} ?
\bigcirc a. x + 3y = 16
\bigcirc b. 2x + y = 12
\bigcirc c. -7x - 5y = -48
\bigcirc d. 7x - 5y = 48

Explanation:

Step1: Find slope of $\overrightarrow{AB}$

Slope formula: $m_{AB}=\frac{y_B - y_A}{x_B - x_A}=\frac{4 - (-1)}{4 - (-3)}=\frac{5}{7}$

Step2: Find slope of $\overrightarrow{BC}$

Perpendicular slopes: $m_{BC}=-\frac{1}{m_{AB}}=-\frac{7}{5}$? No, correction: Perpendicular slope product = -1, so $m_{BC}=-\frac{7}{5}$ is wrong. Wait, correct: $m_{AB} \times m_{BC}=-1$, so $m_{BC}=-\frac{7}{5}$? No, $\frac{5}{7} \times m_{BC}=-1$, so $m_{BC}=-\frac{7}{5}$. Wait no, test options: Use point B(4,4) to check which equation satisfies and has perpendicular slope.

Step2: Test point B in options

For option A: $4 + 3(4)=4+12=16$, holds. Slope of A: $y=-\frac{1}{3}x+\frac{16}{3}$, slope $-\frac{1}{3}$. Wait, recalculate $m_{AB}$: $\frac{4 - (-1)}{4 - (-3)}=\frac{5}{7}$. Wait, $\frac{5}{7} \times (-\frac{7}{5})=-1$, so correct perpendicular slope is $-\frac{7}{5}$. Wait, check option C: $-7(4)-5(4)=-28-20=-48$, holds. Slope of C: $y=-\frac{7}{5}x+\frac{48}{5}$, slope $-\frac{7}{5}$. Wait, $\frac{5}{7} \times (-\frac{7}{5})=-1$, which is perpendicular. Wait no, original calculation: $\overrightarrow{AB}$ vector is (4 - (-3), 4 - (-1))=(7,5). Perpendicular vector has dot product 0: (7,5)·(a,b)=7a+5b=0, so 7a=-5b, so direction vector (5,-7) or (-5,7). Slope is $\frac{-7}{5}$ or $\frac{7}{-5}$. So line through (4,4) with slope $-\frac{7}{5}$: $y-4=-\frac{7}{5}(x-4)$ → $5y-20=-7x+28$ → $7x+5y=48$ → $-7x-5y=-48$, which is option C. Wait, earlier mistake: $\frac{5}{7} \times (-\frac{7}{5})=-1$, correct. So step correction:

Step1: Calculate $\overrightarrow{AB}$ vector

$\overrightarrow{AB}=(4 - (-3), 4 - (-1))=(7,5)$

Step2: Find perpendicular slope

For perpendicular lines, $m_1 \times m_2=-1$, so $m_{BC}=-\frac{7}{5}$

Step3: Write line equation for $\overrightarrow{BC}$

Use point-slope form: $y - 4 = -\frac{7}{5}(x - 4)$
Multiply by 5: $5(y-4)=-7(x-4)$ → $5y-20=-7x+28$
Rearrange: $7x + 5y = 48$ → $-7x -5y = -48$

Step4: Verify with point B

Substitute (4,4) into $-7x-5y=-48$: $-7(4)-5(4)=-28-20=-48$, which is true.

Step1: Compute $\overrightarrow{AB}$ vector

$\overrightarrow{AB}=(4-(-3), 4-(-1))=(7,5)$

Step2: Find perpendicular slope

Since $\overrightarrow{AB} \perp \overrightarrow{BC}$, $m_{AB} \times m_{BC}=-1$. $m_{AB}=\frac{5}{7}$, so $m_{BC}=-\frac{7}{5}$

Step3: Derive line equation

Use point-slope form with $B(4,4)$:
$y-4 = -\frac{7}{5}(x-4)$
Multiply by 5: $5y-20=-7x+28$
Rearrange: $-7x-5y=-48$

Step4: Verify equation

Substitute $x=4,y=4$: $-7(4)-5(4)=-48$, holds true.

Answer:

A. $x + 3y = 16$