JEE Novice - (29)

Geometry Level 3

$\Large {7x-11y+3=0 \\ 4x+3y-9=0 \\ 13x+py-48=0}$

Find the value of $p$ for which the above lines are concurrent.

This question is a part of JEE Novices .

The answer is 26.

1 solution

Rohit Ner
May 28, 2015

From the first equation: $x=\frac { 11y-3 }{ 7 }$ .
Substituting the value in the second equation, we get
$4\left( \frac { 11y-3 }{ 7 } \right) +3y-9=0\\ 44y-12+21y-63=0\\ 65y=75\\ y=\frac { 75 }{ 65 }$
From the third equation, we get
$13\left( \frac { 11y-3 }{ 7 } \right) +py-48=0\\ 143y-39+7py=336\\ y\left( 143+7p \right) =375\\ 143+7p=\frac { 375 }{ y } =\frac { 375 }{ \frac { 75 }{ 65 } } =325\\ 7p=182\\ p=\huge\boxed { 26 }$

Let me elaborate your solution :

Since we know the equations of 2 lines , we can find their intersection point say. $A$ , by solving the equations simultaneously. Now this point $A$ satisfies the equation of the 3rd line having the variable $p$ since all the three lines are concurrent. So by substituting the coordinates of point $A$ in the equation of the 3rd line , we get a linear equation in $p$ and in this way we find the value of $p$ .

Nihar Mahajan - 6 years ago

Yaah,it makes a perfect solution now!

Rohit Ner - 6 years ago

Nicely explained. :)

For the solving part done in Rohit Ner's solution, I'd guess that cross multiplication is much more efficient than substitution in solving the first two equations.

Prasun Biswas - 6 years ago

Thanks!!! :)

Nihar Mahajan - 6 years ago

We can also do it the following way: Determinant of the coefficients in the given equations should be equal to zero if the lines are concurrent. By expanding the determinant and solving.

We get p = 26.

Sameer Kolhar - 6 years ago

JEE Novice - (29)

The answer is 26.

1 solution

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