What is the equation of a line that passes through the point (10, 5) and is perpendicular to the line whose equation is y=54xβ2 ? Enter your answer in the box.
Accepted Solution
A:
y = 5/4x - 2....the slope here is 5/4. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal slope, u take ur original slope, flip it, and change the sign. So we have 5/4.....flip it an it becomes 4/5...change the sign and it becomes - 4/5. So ur perpendicular line will have a slope of -4/5.
y = mx + b slope(m) = -4/5 (10,5)...x = 10 and y = 5 now sub into the formula and find b, the y int 5 = -4/5(10) + b 5 = -40/5 + b 5 = -8 + b 5 + 8 = b 13 = b so ur perpendicular equation is : y = -4/5x + 13 <===