How do you write the slope-intercept form of an equation of the line that passes through the given point and is perpendicular to the equation given (-4,-3), #4x + y=7 #?

Question

How do you write the slope-intercept form of an equation of the line that passes through the given point and is perpendicular to the equation given (-4,-3), #4x + y=7  #?

Eli Assouline · Accepted Answer

Line 1: y = -4x + 7

Slope of the Line 2 that is perpendicular to Line 1 is 4

y2 = 4x + b. Find b.

Write Line 2 passes at point (-4, -3):

-3 = -16 + b -> b = 16 - 3 = 13

Equation of Line 2: y2 = 4x + 13.

Jace Anderson · Answer

undefined To find the slope-intercept form of the equation of a line that passes through a given point and is perpendicular to another line, we first need to determine the slope of the given line.

The given equation is $4x + y = 7$. To rewrite it in slope-intercept form ($y = mx + b$), we isolate $y$ by subtracting $4x$ from both sides:

$$ y = -4x + 7 $$

Now, the slope of the given line is -4.

Since the line we want to find is perpendicular to the given line, its slope will be the negative reciprocal of -4, which is $ \frac{-1}{-4} = \frac{1}{4} $.

Now, we have the slope of the new line and a point it passes through (-4, -3). We can use the point-slope form of a linear equation:

$$ y - y_1 = m(x - x_1) $$

Substituting the given point $(-4, -3)$ and the slope $m = \frac{1}{4}$:

$$ y - (-3) = \frac{1}{4}(x - (-4)) $$

$$ y + 3 = \frac{1}{4}(x + 4) $$

Now, let's simplify:

$$ y + 3 = \frac{1}{4}x + 1 $$

$$ y = \frac{1}{4}x - 2 $$

So, the slope-intercept form of the equation of the line passing through (-4, -3) and perpendicular to $4x + y = 7$ is $y = \frac{1}{4}x - 2$.