Select the TWO lines from this list that are perpendicular.
4x−3y=−6
4 x minus 3 y is equal to negative 6
y−2=34(x+1)
y minus 2 is equal to 3 fourths times open paren x plus 1 close paren
y=−34x+12
y is equal to negative 3 fourths x plus 12
5x−3y=−3
To determine if two lines are perpendicular, we need to look at their slopes. If the product of the slopes of two lines is -1, then the lines are perpendicular to each other.
Let's find the slopes of the given lines by putting them into slope-intercept form, y = mx + b, where m is the slope.
1. For the first line 4x − 3y = −6, we solve for y:
3y = 4x + 6
y = (4/3)x + 2
The slope of this line (m1) is 4/3.
2. For the second line y − 2 = 3/4(x + 1), the equation is already in slope-intercept form, so the slope (m2) is 3/4.
3. For the third line y = −3/4x + 12, the equation is in slope-intercept form, and the slope (m3) is −3/4.
4. For the fourth line 5x − 3y = −3, we solve for y:
3y = 5x + 3
y = (5/3)x + 1
The slope of this line (m4) is 5/3.
Now we need to check the pairs of slopes to find which pair of lines are perpendicular. The slopes of lines perpendicular to each other would multiply to give -1.
Multiplying the slope of the first line (m1 = 4/3) and the slope of the third line (m3 = −3/4) gives us:
(4/3) * (−3/4) = −1
This means that the lines represented by the equations 4x − 3y = −6 and y = −3/4x + 12 are perpendicular to each other. The other pairs of lines do not have slopes that multiply to give -1, so they are not perpendicular.