Find the equation of a straight line which makes acute angle with positive direction of x-axis, passes through point(-5,0) and is at a perpendicular distance of 3 units from origin.
Answers
Answer:
Step-by-step explanation:
Step-by-step explanation
4x - 3y + 20 = 0 is the equation of a straight line which makes acute angle with positive direction of x-axis, passes through point(-5,0) and is at a perpendicular distance of 3 units from origin
Step-by-step explanation:
y = mx + c is equation of line
it passes through (- 5 , 0)
=> 0 = -5m + c
=> c = 5m
y = mx + 5m
perpendicular line
y = -x/m + c
passes through (0, 0)
=> 0 = 0 + c
=> c = 0
=> y = -x/m
=> my = -x
=> x + my = 0
=> x + m(mx + 5m) = 0
=> x = -5m/(m² + 1)
y = 5/(m² + 1)
Its at distance of 3 units from origin
=> (y - 0)² + (x - 0)² = 3²
=> x² + y² = 9
=>( -5m/(m² + 1))² + (5/(m² + 1))² = 9
=> 25 (m² + 1) = 9(m² + 1)²
=> m² + 1 = 25/9
=> m² = 16/9
=> m = ± 4/3
angle is acute with with positive direction of x-axis
=> m = 4/3
y = mx + 5m
y = 4x/3 + 5(4/3)
=> 3y = 4x + 20
=> 4x - 3y + 20 = 0
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