Question

What is the equation of a line that goes through the point (10, -3) and is perpendicular to the line y = 5x + 2

Answer 1

Step-by-step explanation:

Given a line already, the perpendicular line's slope can be found from the fact the multiplication of the two lines' slopes must equal -1.

m1*m2 = -1 --> m1 = 7/5, so m2 = -5/7

To have the second line go through the point (2,-6). we need to solve for b

y = m2x + b

-6 = -5/7(2) + b

b = -6 + 10/7 = -32/7

the perpendicular line which goes through the requested point is...

y = -5x/7 - 32/7 = -(5x + 32)/7

Answer 2

Solution -

We have already an equation of a line, i.e.,

• y = 5x + 2

Therefore, according to intercept form it's slope will be 5.

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Now, looking towards the second line that is perpendicular to the first line. It goes through a point A(10, -3).

• Let another point on that line B(x, y).

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We have to find the slope of the AB.

To find slope, we use

$\: \: \: \: \: \: \: \: \dag\bf\: \: \: {\red{m = \dfrac{(y_2 - y_1)}{(x_2 - x_1)}}}$

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Putting values

$\tt\dashrightarrow{m_2 = \dfrac{y - (-3)}{x - 10}}$

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$\tt\dashrightarrow{m_2 = \dfrac{y + 3}{x - 10}}$

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Since, when two lines are perpendicular to each other, multiplication of their slopes will be -1.

We have,

• $\sf{m_1 = 5}$
• $\sf{m_2 = \dfrac{y + 3}{x - 10}}$

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According to perpendicularity

$\tt:\implies\: \: \: \: \: \: \: \: {m_1 \times m_2 = -1}$

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$\tt:\implies\: \: \: \: \: \: \: \: {5 \times \dfrac{y + 3}{x - 10} = -1}$

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$\tt:\implies\: \: \: \: \: \: \: \: {5(y + 3) = -1(x - 10)}$

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$\tt:\implies\: \: \: \: \: \: \: \: {5y + 15 = -x + 10}$

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$\tt:\implies\: \: \: \: \: \: \: \: {x + 5y + 15 - 10 = 0}$

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$\bf:\implies\: \: \: \: \: \: \: \: {x + 5y + 5 = 0}$

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Hence,

• Required equation of the line is x + 5y + 5 = 0.

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