If possible, can anyone help with these two? --sorry, im dumb ;-;-- THANKS!
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The equation of the line perpendicular to y = 3x + 6 and containing the point (-9,-5) is
<em><u>Solution:</u></em>
Given that line perpendicular to y = 3x + 6 and containing the point (-9, -5)
We have to find the equation of line
<em><u>The slope intercept form is given as:</u></em>
y = mx + c ------ eqn 1
Where "m" is the slope of line and "c" is the y - intercept
<em><u>Let us first find the slope of line</u></em>
The given equation of line is y = 3x + 6
On comparing the given equation of line y = 3x + 6 with eqn 1, we get,
m = 3
Thus the slope of given equation of line is 3
We know that <em>product of slopes of given line and slope of line perpendicular to given line is equal to -1</em>
Slope of given line slope of line perpendicular to given line = -1
Let us now find the equation of line with slope and containing the point (-9, -5)
Substitute and (x, y) = (-9, -5) in eqn 1
<em><u>Thus the required equation of line is:</u></em>
Substitute and c = -8 in eqn 1
Thus the required equation of line perpendicular to given line is found