Question

If the coordinates of points A and B are (-2,-2) and (2,-4) respectively , find the coordinates of the point P such that AP = 3/7 AB , where P lies on the line segment AB​

Answer 1

Step-by-step explanation:

$ap = \frac{3}{7} ab$

$consider \: the \: point \: p(x.y)$

$by \: using \: distance \: formula$

$ap = \frac{3}{7} ab$

$\huge\sf\green{Answer}$

$\sqrt{ {(y2 - y1) }^{2} + {(x2 - x1)}^{2} } = \frac{3}{7} \times \sqrt{ {(y2 - y1)}^{2} + {(x2 - x1)}^{2} }$

$\sqrt{ {(y2 - ( - 2)}^{2} + {(x2 - ( - 2)}^{2} } = \frac{ 3}{7} \times \sqrt{ { - 4 - ( - 2)}^{2} + {(2 - ( - 2)}^{2}) }$

${(y2 + 4)}^{2} + {(x2 + 4)}^{2} = \frac{9}{49} \times { { (- 4 + 2)}^{2} + {(2 - ( - 2)}^{2}) }$

$49( {y 2+ 4)}^{2} + {(x2 + 4)}^{2} = 9 \times {( - 2)}^{2} + {(4)}^{2} )$

$49( {y}^{2} + 16 + 8y) + {x}^{2} + 8x + 16 = 9 \times 4 + 16$

$49{y}^{2} + 784 + 392y + {x}^{2} + 8x + 16 = 52$

${49y}^{2} + {x}^{2} + 8x + 392y + 800 = 16$

${49y}^{2} + {x}^{2} + 8x + 392y + 800 - 16 = 0$

${49y}^{2} + {x}^{2} + 8x + 392y + 784 = 0$