Write an equation in slope intercept form of the line that is perpendicular to the given line and that passes through the given point
3x-4y=8 ; (-6,5)
y-5 = -4/3 (x+6)
now just convert that to y = mx+b
When I did it I got y=3/4x+9 1/2... Did u get that also ?
y - 5 = -4/3 (x+6)
y - 5 = -4/3 x - 8
y = -4/3 x - 3
the slope has to be -4/3, since the original line had slope 3/4 and you want a perpendicular line.
Your line is parallel to it, not perpendicular.
And, as long as you're doing algebra now, lose the mixed numbers. They're just a nuisance. 19/2 is much more useful in calculations than 9 1/2.
To find the equation of a line that is perpendicular to the given line and passes through the given point, we can follow these steps:
1. First, let's rearrange the given equation into slope-intercept form (y = mx + b), where m represents the slope and b represents the y-intercept.
3x - 4y = 8
Subtract 3x from both sides:
-4y = -3x + 8
Divide both sides by -4 to solve for y:
y = (3/4)x - 2
2. Recall that perpendicular lines have slopes that are negative reciprocals of each other. The negative reciprocal of the slope (3/4) is -4/3.
3. Now, we have the slope (-4/3) of the perpendicular line and the given point (-6, 5). We can plug these values into the point-slope formula (y - y1 = m(x - x1)).
y - 5 = (-4/3)(x - (-6))
Simplify:
y - 5 = (-4/3)(x + 6)
4. To convert this equation to slope-intercept form, let's distribute (-4/3) to (x + 6):
y - 5 = (-4/3)x - 8
5. Finally, isolate y by adding 5 to both sides:
y = (-4/3)x - 3
Therefore, the equation in slope-intercept form of the line perpendicular to 3x - 4y = 8 and passing through the point (-6, 5) is y = (-4/3)x - 3.