- b.First, convert the standard linear equation into a slope-y
intercept equation. Isolate the yvariable: 2y= –1x+ 4. Divide both
sides by 2: y= –^12 x+ 2. A line that perpendicularly intercepts this
line on the y-axis has a negative reciprocal slope but has the same y
intercept value: y= 2x+ 2.
- d.Change ^12 x+ ^14 y= ^18 into slope-intercept form by subtracting ^12 x
from both sides and multiplying everything by 4 (to get rid of the
^14 next to y). The slope-intercept form is y= –2x+ ^12 , which means
that the slope is –2. Since perpendicular lines have negative reciprocal
slopes, all lines that have a slope of ^12 will be perpendicular to ^12 x+ ^14 y
= ^18 . When changed into slope-intercept form, choice dbecomes
y= –^12 x+ 3 which has a slope of –^12 , so this will not be perpendicular
to the original equation.
- c.Change 2x+ 3y= 6 into slope-intercept form by subtracting 2x
from both sides and dividing everything by 3 (to get rid of the 3
next to y). The slope-intercept form is y= –^23 x+ 2, which means
that the slope is –^23 . Since parallel lines have equal slopes, all lines
that have a slope of –^23 .will be parallel to 2x+ 3y= 6. Choice c,
y= –^23 x– 5 is the only equation that has a slope of –^23 .
- (–3,–4).Sub “1 + y” in for x in the other equation and solve fory:
10(1 + y) – 9y= 6
10 + 10y– 9y= 6
1 y= 6 – 10, so y= –4
Sub y= – 4 in for yand solve for x:
x= 1 + (– 4), x= –3
- (0,4)Set the two equations equal to one another and then solve for x:
(^23 )x+ 4 = –1x+ 4
(^23 )x+ 1x= 4 – 4
(^53 )x= 0
x= 0
Sub x= 0 in for xand solve for y:
y= –1(0) + 4, y= 4
501 Geometry Questions
