Pre calculas help!?
1.x intercept= -5 y-intercept= -5
2. through (-1,3) and perpendicular to the line through (0,1) and (4,3)
3. y-intercept 3 and perependicular to 2x-y+6=0
4. find the real number k such that 3,-2 is on the line kx-2y+7=0
Favorite Answer
y-intercept = (0, -5)
slope = m = rise/run = (y2 – y1) / (x2 – x1) = -5/5 = -1
Put the line in y = mx + b format.
y – y1 = m(x – x1)
y – 0 = -1(x + 5)
y = -x – 5
2. slope = (3 – 1)/(4 – 0) = 2/4 = 1/2
The perpendicular line’s slope is the negative reciprocal of that, i.e., -2.
y – 3 = -2(x + 1)
y = -2x + 1
3. y-intercept (0, 3)
-y = -2x – 6
y = 2x + 6
m = 2 so the negative reciprocal is -1/2
y – 3 = -(1/2)(x – 0)
y = -x/2 + 3
4. Plug in x and y and solve for k.
k(3) – 2(-2) + 7 = 0
3k + 11 = 0
k = -11/3
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