# Find the slope ‐ intercept equation of the line with the following properties: Perpendicular to the line x - 4y = 2 containing the point (5,2)

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The required line is perpendicular to the line x - 4y = 2. This is the case when the product of the slope of the line and that of x - 4y = 2 is equal to -1.

x - 4y = 2 can be written as y = x/4 - 2/4 in slope intercept form. The slope of the required line is -4.

As it has the point (5, 2) its equation is:

(y - 2)/(x - 5) = -4

=> y - 2 = -4x + 20

=> y = -4x + 22

**The equation of the required line in slope-intercept form is y = -4x + 22**

its simple

the slope of reqd line will be -4. and contains the pt (5,2)

(y-2)= -4(x- 5)

y+4x=22 divide by 22

y/22 + x/5.5= 1

Perpendicular to the line x - 4y = 2 containing the point (5,2)

Lets take your standard form equation and change it to slope-intercept to readily identify the slope:

4y = x - 2

y = (1/4)x - (1/2)

The slope is 1/4.

A perpendicular line's slope will be the negative reciprocal of this one so -4.

Now we have a point and a slope, lets first write out the perpendicular line in point-slope form.

y - 2 = -4(x - 5)

Distribute the -4 and solve for y to put this into slope-intercept form.