Question

What is the equation of the line described below written in slope-intercept form? the line passing through point (4, -1) and perpendicular to the line whose equation is 2x - y - 7 = 0 y = -x + 1 y = x - 1 y = 2x + 1

Answer 1

Answer:

The equation of the perpendicular line $y = -\frac{1}{2} x + 1$

Step-by-step explanation:

Given: The line passing through point (4, -1) and perpendicular to the line whose equation is 2x - y - 7 = 0.

The equation of line given in standard form.

We need to write in slope-intercept form.

The general form of slope-intercept form is y = mx + b, where "m" is the slope and b is the y-intercept.

2x - y - 7 = 0

Step 1:

Add 7 on both sides, we get

2x - y - 7 + 7 = 0 + 7   [Addition of equality]

2x - y = 7

Step 2:

Subtract 2x from both sides, we get

2x - 2x - y = -2x + 7

-y = -2x + 7

Step 3:

We have negative y, we need to make it positive, so we need to multiply both sides by -1.

-1(-y) = -1(-2x + 7)

y = 2x - 7

Here slope (m) = 2 and y-intercept (b) = -7

Now we have to find the perpendicular line.

The slope of the perpendicular line is negative reciprocal of the slope of the given line.

So, m = -1/2

Now we have to find y-intercept.

We are given a point (4, -1)

Now plug in x = 4 and y = -1 in y = -1/2x + b

-1 = -1/2(4) + b

-1 = -2 + b

b = -1 +2

b = 1

So, the equation of the perpendicular line $y = -\frac{1}{2} x + 1$