Solve for t 2t^2-14t+3=3
1 answer:
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<u>Answer:</u>
The equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.
<u>Solution:</u>
Given, line equation is y = x – 1 ⇒ x – y – 1 = 0. And a point is (-3, -2)
We have to find the line equation which is perpendicular to above given line and passing through the given point.
Now, let us find the slope of the given line equation.
We know that, <em>product of slopes of perpendicular lines is -1. </em>
So, 1 slope of perpendicular line = -1
slope of perpendicular line = -1
Now let us write point slope form for our required line.
y – (-2) = -1(x – (-3))
y + 2 = -1(x + 3)
y + 2 = -x – 3
x + y + 2 + 3 = 0
x + y + 5 = 0
y = -x -5
Hence the equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.