A person standing at the crossing at two straight laths represented the equations 2x–3x–4=0 and 3x–4y–5=0 wants to reach a path represented by 6x–7y+8=0 in least time. Find the equations of path he should follow
SARDARshubham:
is the first equation is 2x-3x-4 =0
Yes
i think there should be y in this equation
ya there is x and y both
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we are required to find the equation of the line perpendicular to the line whose equation is
6x - 7 y + 8=0
and that passes through the point of intersection of the two lines whose equations are 2x - 3y + 4 =0 and 3 x + 4y - 5 = 0 .
The point of intersection of the two lines: 2x - 3y + 4 =0 and 3 x + 4y -5 =0
i.e 4 ×(2x - 3y + 4 =0) and 3×( 3 x + 4y -5 =0)
⇒8x- 12y+16= 0 and 9x+12y-15= 0
⇒8x-12y+16+ 9x+ 12y- 15= 0
⇒17x- 31y= 0
⇒x=31/17 ⇒2(31/7)-3y + 4=0 ⇒y= -2/17
now the slope of the given line 6x-7y+8=0
comparing it to ax + by +c=0
whose slope is -b/a is given by -7/6
⇒the slope of the required line = -7/6
⇒the equation of the required line is y - y°=m(x - x°)
⇒y - (-2/17) =(-7/6){x - (31/17)}
⇒y + 2/7 =-7x/6 + 31×7/17×6
⇒119x + 102x =205
and that passes through the point of intersection of the two lines whose equations are 2x - 3y + 4 =0 and 3 x + 4y - 5 = 0 .
The point of intersection of the two lines: 2x - 3y + 4 =0 and 3 x + 4y -5 =0
i.e 4 ×(2x - 3y + 4 =0) and 3×( 3 x + 4y -5 =0)
⇒8x- 12y+16= 0 and 9x+12y-15= 0
⇒8x-12y+16+ 9x+ 12y- 15= 0
⇒17x- 31y= 0
⇒x=31/17 ⇒2(31/7)-3y + 4=0 ⇒y= -2/17
now the slope of the given line 6x-7y+8=0
comparing it to ax + by +c=0
whose slope is -b/a is given by -7/6
⇒the slope of the required line = -7/6
⇒the equation of the required line is y - y°=m(x - x°)
⇒y - (-2/17) =(-7/6){x - (31/17)}
⇒y + 2/7 =-7x/6 + 31×7/17×6
⇒119x + 102x =205
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