find equation of line passing through the intersection point 2x-y=14,2x+y=10 and perpendicular line is 3x-y+6=0.
Answers
Answer:
Step 1
Solve the system of equations for the intersection point. If it is different than A(3, 2), then we can use it to determine the slope of the line:
3x - y - 6 = 0
3x - 6 = y
y = 3x - 6 Equation 1
2x + 1 - 14 = 0
2x = -1 + 14 = 13
x = 13/2 Equation 2
Substitute the value of x from Equation 2 into Equation 1, and solve for y.
y = 3x - 6
y = 3(13/2 ) - 6
y = 39/2 - 12/2 = 27/2
y = 27/2
So, we have the intersection point B(13/2, 27/2). We can use this and the given point A(3, 2) to determine the slope, m, as follows:
m = (yB - yA)/(xB - xA) = (27/2 - 2)/(13/2 - 3) = (23/2)/(7/2)
m = 23/7
Step 2
Use the Point-Slope formula to write the equation of the line passing through points A and B.
y - yA = m(x - xA)
y - 2 = 23/7(x - 3)
y - 2 = (23/7)x - 69/7
y = (23/7)x - 69/7 +2
y = (23/7)x - 69/7 + 14/7
y = (23/7)x - 55/7