How do you write an equation in standard form for a line passes through (2, –3) and is perpendicular to y = 4x + 7?

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How do you write an equation in standard form for a line passes through (2, –3) and is perpendicular to y = 4x + 7?

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Answer 1 · 2015-06-06T00:30:58

Answer: the equation of the line is $y = - \frac{1}{4} x - \frac{5}{2}$ $y=-1/4x -5/2$

Explanation: We have two lines: $L_{1}$ $L_1$ defined by the equation $y = a x + b$ $y=ax+b$ and $L_{2}$ $L_2$ defined by the equation $y = 4 x + 7$ $y=4x+7$ $\Rightarrow$ $=>$ The slope of the line $L_{2}$ $L_2$ is $4$ $4$ .

We know that, If $L_{1}$ $L_1$ is perpendicular to $L_{2}$ $L_2$ , then the slope of $L_{1}$ $L_1$ is the inverse reciprocal of the slope of $L_{2}$ $L_2$ . Therefore, the slope of $L_{1}$ $L_1$ is $- \frac{1}{4}$ $-1/4$ .

So, the equation of $L_{1}$ $L_1$ is $- \frac{1}{4} x + b$ $-1/4x + b$

The line $L_{1}$ $L_1$ contains the point $A (2, - 3)$ $A(2,-3)$ , therefore we have the following equation: $- 3 = - \frac{1}{4} \cdot 2 + b$ $-3=-1/4*2+b$ $\Rightarrow$ $=>$ $b = - \frac{5}{2}$ $b=-5/2$

Therefore, the equation of $l_{1}$ $l_1$ is $y = - \frac{1}{4} x - \frac{5}{2}$ $y=-1/4x -5/2$

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How do you write an equation in standard form for a line passes through (2, –3) and is perpendicular to y = 4x + 7?

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