write the slope intercept form of the equation of the line described. then find the distance between the point and the line. -through (-4,0) perp. to y=3x
The line is perpendicular to y=3x, which means it has gradient\[m=-\frac{1}{3}\]Using the point-gradient formula, we have\[y-0=-\frac{1}{3}(x-(-4)) \rightarrow y = -\frac{1}{3}x+\frac{4}{3}\]
brb
okay :)
To find the distance between y=3x and (-4,0), you have to understand, when we say 'distance' in mathematics without any other qualification, we're talking about the *shortest* distance between two objects. In a Euclidean plane (like you have here), the shortest distance between a point and a line is the distance along a line that is perpendicular to the original line. I'll upload a pic.
The blue line is the original equation, y=3x, and the red is the perpendicular one we've found. Since we have the equation of the line perpendicular to y=3x, the distance from (-4,0) to y=3x is just the distance between the point (-4,0) and the point of intersection.
The equation of the perpendicular line should be\[y=-\frac{1}{3}x-\frac{4}{3}\]It's what I have on the plot, but didn't punch it in properly above (hate this thing).
The theory is still the same. Distance between (-4,0) and point of intersection is needed, which is given on the attached file.
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