Question

A straight line has slope

\frac{5}{12}

and it passes through the point

P (3, - \frac{7}{2})

. Find the co-ordinates of a point

Q

on this line which is at a distance of

\frac{13}{2}

unit from

P

Accepted Answer

Given the slope of a line in $\frac{5}{12}$ and it passes through $(3, \frac{- 7}{2})$

$∴$ By point slope form we have equation of line is given by

$y - (\frac{- 7}{2}) = \frac{5}{12} (x - 3)$

$\Rightarrow y \frac{+ 7}{2} = \frac{5}{12} (x - 3)$

$\Rightarrow \frac{2 y + 7}{2} = \frac{5 x - 15}{12}$

$\Rightarrow 12 y + 42 = 5 x - 15$

$5 x - 12 y - 57 = 0 . . . . (1)$

Let coordinates of $Q$ be $(x_{1}, y_{1})$ and it lies on line given by $(1)$

So, we have

$5 x_{1} - 12 y_{1} - 57 = 0 . . . . (2)$

Given that $P Q$ distance $= \frac{13}{2}$

$∴ \sqrt{{(3 - x_{1})}^{2} + {(\frac{- 7}{2} - y_{1})}^{2}} = \frac{13}{2}$

Distance between $(x_{1}, y_{1}) & (x_{2}, y_{2})$ is given by $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

$\Rightarrow$ Squaring both sides

${(3 - x_{1})}^{2} + {(\frac{7}{2} + y_{1})}^{2} = \frac{169}{4}$

$\Rightarrow 9 + x_{1}^{2} - 6 x_{1} + \frac{49}{4} + y_{1}^{2} + 7 y_{1} = \frac{169}{4}$

$\Rightarrow x_{1}^{2} + y_{1}^{2} - 6 x_{1} + 7 y_{1} = \frac{169}{4} - \frac{49}{4} - 9$

$\Rightarrow x_{1}^{2} + y_{1}^{2} - 6 x_{1} + 7 y_{1} = 21 . . . . (3)$

From $(2)$ we have

$x_{1} = \frac{12 y_{1} + 57}{5}$

Putting the value of $x_{1}$ is $(3)$ we get

${(\frac{12 y_{1} + 57}{5})}^{2} + y_{1}^{2} - 6 (\frac{12 y_{1} + 57}{5}) + 7 y_{1} = 21$

$\Rightarrow \frac{144 y_{1}^{2} + 3249 + 1369 y_{1} + y_{1}^{2}}{25} - (\frac{72 y_{1} + 342}{5}) + 7 y_{1} = 21$

$\Rightarrow 144 y_{1}^{2} + 3249 + 1368 y_{1} + 25 y_{1}^{2} + 360 y_{1} - 1710 + 175 y_{1} = 525$

$\Rightarrow 169 y_{1}^{2} + 1183 y_{1} + 1014 = 0$

$\Rightarrow 169 (y_{1}^{2} + 7 y_{1} + 6) = 0$

$\Rightarrow y_{1}^{2} + 6 y_{1} + y_{1} + 6 = 0$

$\Rightarrow (y_{1} + 6) (y_{1} + 1) = 0 \Rightarrow y_{1} = - 1 or - 6$

When $y_{1} = - 1$

$x_{1} = \frac{- 12 + 57}{5} = \frac{45}{5} = 9$

Where $y_{1} = - 6$

$x_{1} = \frac{12 (- 6) + 57}{5} = \frac{- 72 + 57}{5} = \frac{- 15}{5} = - 3$

Hence the coordinate of $P$ are $(9 - 1)$ or $(- 3, - 6)$ .

A straight line has slope $\frac{5}{12}$ and it passes through the point $P (3, - \frac{7}{2})$ . Find the co-ordinates of a point $Q$ on this line which is at a distance of $\frac{13}{2}$ unit from $P$

Important Questions on Straight Line

Learn and Prepare for any exam you want !

A straight line has slope 512 and it passes through the point P3, -72. Find the co-ordinates of a point Q on this line which is at a distance of 132 unit from P

Important Questions on Straight Line

Learn and Prepare for any exam you want !

A straight line has slope $\frac{5}{12}$ and it passes through the point $P (3, - \frac{7}{2})$ . Find the co-ordinates of a point $Q$ on this line which is at a distance of $\frac{13}{2}$ unit from $P$