Question

Find the equation of the line which passes through the point of intersection of the lines y+2x=8 and 2y-x=6 and the point (4,3)

Answer 1

Answer:

Given,

$\bf \: y + 2x = 8 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \implies \: 2x + y - 8 = 0 \: \: \: \: - - - - (1) \\ \\ and \\ \\ 2y - x = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \implies \: x - 2y + 6 = 0 \: \: \: \: - - - - (2)$

We have to find :

A linear equation which is passing through the intersection point of eqn.(1) and eqn.(2)

So our equation is :

$\bf2x + y - 8 + k(x - 2y + 6) = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \implies \: 2x + y - 8 + kx - 2ky + 6k = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \implies \: (2 + k)x + (1 - 2k)y - 8 + 6k = 0 \: \: \: \: - - - (3)$

Eqn.(3) is also passing through the point (4,3)

so (4,3) point will satisfy the equation.

Now,

$(2 + k)x + (1 - 2k)y - 8 + 6k = 0 \: \: \\ \bf \implies \: (2 + k) \times 4 + (1 - 2k) \times 3 - 8 + 6k = 0 \\ \bf \implies \: 8 + 4k + 3 - 6k - 8 + 6k = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \implies \: 4k + 3 = 0 \\ \bf \implies \: k = - \frac{3}{4}$

put the value of k in eqn.(3)

$(2 - \frac{3}{4} )x + (1 + 2 \times \frac{3}{4} )y - 8 - 6 \times \frac{3}{4} = 0 \\ \\ \bf \implies \: \frac{5x}{4} + \frac{5y}{2} - \frac{25}{2} = 0 \\ { \bf \implies \: 5x + 10y - 50 = 0} \\ \green{ \boxed{\bf \implies \: x + 2y - 10 = 0}}$