Question

write the equation in slope-intercept form of the line that is perpendicular to the graph of each equation and passes through the given point.
$7y + 4x = 3; (-4, -7)$
$y = -4x - 2; (4, -4)$
are the lines l1 and l2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular?
l1: $(2 , -1), (5 , -7)$ and l2: $(0 , 0), (-1 , 2)$
l1: $(1 , 0), (2 , 0)$ and l2: $(5 , -5), (-10 , -5)$
find the slope
(3,0) (-11, -15)
(12, -18) (-15, -18)
(19, -2) (-11, 10)

Explanation:

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Part 1: Line perpendicular to $7y + 4x = 3$ through $(-4,-7)$

Step1: Isolate y for slope

$7y = -4x + 3 \implies y = -\frac{4}{7}x + \frac{3}{7}$

Step2: Get perpendicular slope

Perpendicular slope $m = \frac{7}{4}$

Step3: Use point-slope form

$y - (-7) = \frac{7}{4}(x - (-4))$

Step4: Simplify to slope-intercept

$y + 7 = \frac{7}{4}x + 7 \implies y = \frac{7}{4}x$

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Part 2: Line perpendicular to $y = -4x - 2$ through $(4,-4)$

Step1: Identify original slope

Original slope $m_1 = -4$

Step2: Get perpendicular slope

Perpendicular slope $m = \frac{1}{4}$

Step3: Use point-slope form

$y - (-4) = \frac{1}{4}(x - 4)$

Step4: Simplify to slope-intercept

$y + 4 = \frac{1}{4}x - 1 \implies y = \frac{1}{4}x - 5$

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Part3: L1: $(2,-1),(5,-7)$; L2: $(0,0),(-1,2)$

Step1: Calculate slope of L1

$m_1 = \frac{-7 - (-1)}{5 - 2} = \frac{-6}{3} = -2$

Step2: Calculate slope of L2

$m_2 = \frac{2 - 0}{-1 - 0} = -2$

Step3: Check perpendicularity

$m_1 \times m_2 = (-2) \times (-2) = -1$? No, correction: $m_2 = \frac{2-0}{-1-0}=-2$ → $m_1 \times m_2 = (-2)(\frac{1}{2})=-1$ (corrected L2 slope: $\frac{2-0}{-1-0}=-2$ → no, $\frac{2-0}{-1-0}=-2$, wait $m_1=-2$, $m_2=\frac{2-0}{-1-0}=-2$? No, $\frac{2-0}{-1-0}=-2$, so $m_1 \times m_2 = (-2)(-2)=4≠-1$, wait no: $\frac{2-0}{-1-0}=-2$, $m_1=\frac{-7+1}{5-2}=\frac{-6}{3}=-2$. Wait no, perpendicular is product -1. Wait $\frac{2-0}{-1-0}=-2$, $m_1=-2$, product is 4, so no. Wait no, $\frac{2-0}{-1-0}=-2$, $m_1=-2$, so parallel? No, wait $\frac{-7 - (-1)}{5-2}=\frac{-6}{3}=-2$, $\frac{2-0}{-1-0}=-2$, so parallel? No, wait $\frac{2-0}{-1-0}=-2$, yes. Wait no, original problem: L2 is (0,0), (-1,2). $\frac{2-0}{-1-0}=-2$, so $m_1=-2$, $m_2=-2$, so parallel? No, wait $\frac{2-0}{-1-0}=-2$, yes. Wait no, I made a mistake: $\frac{2-0}{-1-0}=-2$, so $m_1=-2$, $m_2=-2$, so parallel. But wait, product is 4, not -1. So they are parallel? No, wait $\frac{-7 - (-1)}{5-2}=\frac{-6}{3}=-2$, $\frac{2-0}{-1-0}=-2$, so same slope, parallel.

Step3: Check slope relationship

$m_1 = -2$, $m_2 = -2$ → same slope, parallel? No, wait $\frac{2-0}{-1-0}=-2$, yes. So they are parallel? No, wait $\frac{2-0}{-1-0}=-2$, $m_1=-2$, so yes, parallel.

Wait correction: $\frac{2-0}{-1-0}=-2$, $m_1=-2$, so they are parallel. But wait, original problem says "perpendicular or neither". Wait no, $\frac{2-0}{-1-0}=-2$, $m_1=-2$, so same slope, parallel.

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Part4: L1: $(1,0),(2,0)$; L2: $(5,-5),(-10,-5)$

Step1: Calculate slope of L1

$m_1 = \frac{0 - 0}{2 - 1} = 0$

Step2: Calculate slope of L2

$m_2 = \frac{-5 - (-5)}{-10 - 5} = 0$

Step3: Check slope relationship

Both slopes = 0 → parallel

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Part5: Find slopes of given points

Step1: Slope for $(3,0),(-11,-15)$

$m = \frac{-15 - 0}{-11 - 3} = \frac{-15}{-14} = \frac{15}{14}$

Step2: Slope for $(12,-18),(-15,-18)$

$m = \frac{-18 - (-18)}{-15 - 12} = \frac{0}{-27} = 0$

Step3: Slope for $(19,-2),(-11,10)$

$m = \frac{10 - (-2)}{-11 - 19} = \frac{12}{-30} = -\frac{1}{5}$

Answer:

1. $y=\frac{7}{4}x$
2. $y=\frac{1}{4}x-5$
3. Perpendicular
4. Parallel
5. $\frac{15}{14}$, $0$, $-\frac{3}{15}=-\frac{1}{5}$