Two perpendicular lines PQ and QR intersect at (1, -1). I...
Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR
x + 2y - 1= -0
2x + y - 3 = 0
x - 2y - 3 = 0
2x + y - 1 = 0
Correct answer is D
Line PQ : x - 2y + 4 = 0
2y = x + 4 ⟹y=x2+2
Slope = 12
Slope of the perpendicular line QR: −112=−2
Line QR: y=mx+b
y=−2x+b
Point of intersection: (1, -1)
−1=−2(1)+b⟹b=−1+2=1
y=−2x+1⟹y+2x−1=0
QR:2x+y−1=0
Given that S and T are sets of real numbers such that S = {x : 0 ≤ x ≤ 5} and T = {x :...
if p = {-3<x<1} and Q = {-1<x<3}, where x is a real number, find P n Q....
If logax = p, express x in terms of a and p ...
Use the graph of sin (θ) above to estimate the value of θ when sin (θ) = -0.6 for ...
In the diagram above, AB//CD, the bisector of ∠BAC and ∠ACD meet at E. Find the value of &an...