Find the equation of the line that contains the point
P(−3, 3)
and is perpendicular to the graph of
2x + 4y = −3.
I got y=-2x+3/2 but it keeps saying that I'm doing it wrong.
2x+4y=-3 has slope -1/2
your line must then have slope 2, not -2.
y = 2x+9
To find the equation of a line perpendicular to a given line, we need to determine the slope of the given line first. The slope of the given line can be found by rearranging the equation into slope-intercept form (y = mx + b), where m represents the slope.
Given: 2x + 4y = -3
Rearranging the equation, we isolate y:
4y = -2x - 3
y = (-2/4)x - (3/4)
y = (-1/2)x - (3/4)
The slope of the given line is -1/2.
Since we want to find the equation of a line that is perpendicular to this line, we need to find the negative reciprocal of the slope. The negative reciprocal of -1/2 is 2/1 or simply 2.
Now that we have the slope of our desired line, we can use the point-slope form of a linear equation y - y1 = m(x - x1), where (x1, y1) represents the coordinates of the given point P(-3, 3).
Plugging in the values, we get:
y - 3 = 2(x - (-3))
y - 3 = 2(x + 3)
y - 3 = 2x + 6
Now, let's rearrange the equation to get it in slope-intercept form by isolating y:
y = 2x + 6 + 3
y = 2x + 9
Therefore, the correct equation for the line that contains point P(-3,3) and is perpendicular to the graph of 2x + 4y = -3 is y = 2x + 9.