the lones x-2y+6=0 and 2x-y-10+0 intersect at P Without finding the coordinates of prove that the equation of the line - Maths - Straight Lines

Question

the lones x-2y+6=0 and 2x-y-10+0 intersect at P Without finding
the coordinates of prove that the equation of the line through P
and the origin of coordinates is perpendicular to 39x+33y-580=0

Manish Bharti · Accepted Answer

First write the equation of line $39 x + 33 y - 580 = 0$ in the form of $y = - \frac{39}{33} x + \frac{580}{33}$ , now slope of the line is
$m_{1} = - \frac{39}{33} = - \frac{13}{11}$
Slope of line perpendicular to the above line is-
$m_{2} = - \frac{1}{m_{1}} = \frac{11}{13}$
Equation of line passing through the origin and with slope equal to 11/13 is given by-
$y = \frac{11}{13} x o r 11 x - 13 y = 0$
Substituting $y = \frac{11}{13} x$ in $x - 2 y + 6 = 0$ , we get
$x - 2 \times \frac{11}{13} x + 6 = 0 x (\frac{13 - 22}{13}) = - 6 x = \frac{6 \times 13}{9} = \frac{26}{3} \Rightarrow y = \frac{11}{13} \times \frac{26}{3} = \frac{22}{3}$

Substituting $y = \frac{11}{13} x$ in $2 x - y - 10 = 0$ , we get
$2 x - \frac{11}{13} x - 10 = 0 x (\frac{26 - 11}{13}) = 10 x = \frac{10 \times 13}{15} = \frac{26}{3} \Rightarrow y = \frac{11}{13} \times \frac{26}{3} = \frac{22}{3}$
Since the point of intersection of lines $y = \frac{11}{13} x$ , $x - 2 y + 6 = 0$ and $2 x - y - 10 = 0$ is same. The line passes through P and origin is perpendicular to $39 x + 33 y - 580 = 0$ .

the lones x-2y+6=0 and 2x-y-10+0 intersect at P.Without finding the coordinates of prove that the equation of the line through P and the origin of coordinates is perpendicular to 39x+33y-580=0.