Find the locus of the point whose distance from the line AB is only 7 cm.
Answers
9x² + 16y² + 24xy + 42x + 56y = 1176
Step-by-step explanation:
We consider the AB line having equation
3x + 4y + 7 = 0
Let the coordinates of the given point be (p, q)
Then the distance of the point (p, q) from AB line be
= ( | 3p + 4q + 7 | ) /√(3² + 4²) units
= ( | 3p + 4q + 7 | ) / 5 units
According to the question,
( | 3p + 4q + 7 | ) / 5 = 7
or, | 3p + 4q + 7 | = 35
or, (3p + 4q + 7)² = 35²
or, 9p² + 16q² + 49 + 24pq + 42p + 56q = 1225
or, 9p² + 16q² + 24pq + 42p + 56q = 1176
∴ the required locus of the point (p, q) be
9x² + 16y² + 24xy + 42x + 56y = 1176
Find more questions here:
• Find equation of the perpendicular dropped from the point (-1,2) onto the line joining the points (1,4) and (2,3) - https://brainly.in/question/6488655
• Let L be a normal to the parabola y² = 4x. If L passes through the point (9, 6), then L is given by - https://brainly.in/question/9607969