Question

The abscissa of a point A is twice its ordinate
and B = (10,0). Find the co-ordinates of A if
AB = 5 units.

Answer 1

Answer:

hope it helps

mark me brainliest

Answer 2

Step-by-step explanation:

Abscissa = x coordinate

Ordinate = y coordinate

According to the given information:

x = 2y...... (1)

Thus coordinates of point A are (x, y) = (2y, y)

B (10, 0)

Now by distance formula:

$AB = \sqrt{(2y - 10)^{2} + (y - 0) ^{2} } \\ \therefore \: 5 = \sqrt{(2y - 10)^{2} + (y - 0) ^{2} } \\ squiring \: both \: sides \\ {5}^{2} = (2y - 10)^{2} + (y - 0) ^{2} \\ \therefore 25 = 4 {y}^{2} - 40y+ 100 + {y}^{2} \\ \therefore 25 = 5 {y}^{2} - 40y+ 100 \\ \therefore 5 {y}^{2} - 40y+ 100 - 25 = 0 \\ \therefore 5 {y}^{2} - 40y+ 75= 0 \\ \therefore 5 ({y}^{2} - 8y+ 15)= 0 \\ \therefore {y}^{2} - 8y+ 15= 0 \\ \therefore {y}^{2} - 5y - 3y+ 15= 0 \\ \therefore y({y} - 5) - 3(y - 5)= 0 \\ \therefore ({y} - 5)(y - 3)= 0 \\ \therefore ({y} - 5) = 0 \: or \: (y - 3)= 0 \\ \therefore {y} = 5 \: or \: y= 3 \\ when \: y = 5 \implies \: x = 2 \times 5 = 10 \\ when \: y = 3 \implies \: x = 2 \times 3 = 6 \\ thus \: the \: coordinates \: of \: point \: a \: are \\ \: (10, \: 5) \: or \: (6, \: 3)$