Question

15. The line joining A (2, 3) and B 6, -5) meets the
x-axis at P. Write down the y-coordinate of P. Hence

find the ratio AP : PB.
​

Answer 1

Answer:

i). y-coordinate of P is 0

$P = ( \frac{7}{2}, \: 0)$

ii). AP: PB = 3:5

For such kind of questions use Geogebra app. You'll easily find it on Play Store. Also it is helpful for imagining the equations on 2D plane.

Step-by-step explanation:

Given:

Line joining A (2, 3) and B 6, -5) meets the

x-axis at P.

To Find:

i). Y coordinate of P.

ii). ratio of AP:PB.

Solution:

$Line \: joining \: AB \: 's \: equation \: can \: be \: obtained \: by \: Two-Point \: form. \: i.e. \\ \frac{y - y1}{y1 - y2} = \frac{x - x1}{x1 - x2} \\ Here \: (x1, y1) = A(2, 3) \: & \: (x2, y2) = B(6, -5). \\ </p><p>Therefore \\ \frac{(y) - (3)}{(3) - ( - 5)} = \frac{(x) - (2)}{(2) - (6)} \\ \frac{y - 3}{8} = - \frac{ x - 2}{4} \\ (y - 3) = - \frac{8}{4} (x - 2) \\ (y - 3) = - 2(x - 2) \\ (y - 3) = (- 2x + 4) \\ 2x + y - 7 = 0 \\ Hence \: the \: equation \: of \: Line \: AB \: is: \\ 2x + y - 7 = 0 \\ \\ As \: P \: lies \: on \: X \: axis \: y-coordinate of P \: is \: 0\\ Now \: P, \: which \: is \: on \: X \: axis \: lies \: on \: line \: AB. \\ Hence \: say \: P(x,0) \: satisfies \: equation \: of \: line \: AB. \\ </p><p>Therefore \\ 2(x) + (0) - 7 = 0 \\ x = \frac{7}{2} \\ Hence \: x \: coordinate \: of \: P \: is \: \frac{7}{2} = 3.5 \\ P(x,0) = P( \frac{7}{2} ,0) \\ \\ ii). \: \\ By distance formula: \\ d = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }^{2} } \\ Distance \: AP: = d1 = \sqrt{ {(( \frac{7}{2} ) - (2))}^{2} + {((0) - (3))}^{2} } \\ AP: = d1 = \sqrt{ {( \frac{3}{2}) }^{2} + {( - 3)}^{2} } = \sqrt{ { (\frac{9}{4}) + (9)} \\ AP: = d1 = \sqrt{ \frac{45}{4} } \\ \\ Distance \: PB: = d2 = \sqrt{ {(( \frac{7}{2} ) - (6))}^{2} + {((0) - ( - 5))}^{2} } \\ PB: = d2 = \sqrt{ {( - \frac{5}{2} )}^{2} + {(5)}^{2} } \\ \sqrt{( \frac{25}{4} ) + (25)} \\ \sqrt{ \frac{125}{4} } \\ Now, AP : PB = d1 : d2 \\ \frac{AP}{PB} = \frac{ \sqrt{ \frac{45}{4} } }{ \sqrt{ \frac{125}{4} } } \\ \frac{AP}{PB} = \frac{ \sqrt{45} }{ \sqrt{125} } \\\frac{AP}{PB} = \sqrt{ \frac{45}{125} } \\ \frac{AP}{PB} = \sqrt{ \frac{9}{25} } \\ \frac{AP}{PB} = \frac{3}{5} \\ Hence \: AP:PB = 3:5</p><p>$