If (1,p/3) is the mid point …
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Sia ? 6 years, 4 months ago
Since {tex}\left(1,\frac p3\right){/tex}, is the mid point of the line segment joining the points (2,0) and {tex}\left( 0 , \frac { 2 } { 9 } \right){/tex}.
{tex}\therefore \quad \quad \frac { p } { 3 } = \frac { 0 + \frac { 2 } { 9 } } { 2 }{/tex}
{tex} \frac { 2 p } { 3 } = \frac { 2 } { 9 }{/tex}
{tex}p = \frac { 1 } { 3 }{/tex}
Now the given points are (-1, 3p). Let x = -1 , y = 3p
Hence, y = 3 {tex}\times{/tex} {tex}\frac13{/tex} = 1
The equation to prove is: 5x + 3y + 2 = 0
L.H.S.
= 5x + 3y + 2
Put the values of x & y,
= 5(-1) + 3(1) + 2 = -5 + 3 + 2 = -2 + 2 = 0 = R.H.S.
Hence, the line 5x + 3y + 2 = 0 passes through the point (–1, 1) as 5(–1) + 3(1) + 2 = 0.
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