Question

Find the equation of the line passing through the intersection of the lines 3x-4y+1=0 and 5x+y-1=0 and cuts off equal intercepts on both the axis

Answer 1

Question :

Find the equation of the line through the intersection of 3x-4y+1=0 and 5x+y-1=0 which cuts off equal intercepts from the axes.

Solution:

The equation of any line through the point of intersection from the lines

3x-4y+1=0 and 5x+y-1=0

$(3x - 4y + 1) + k(5x + y - 1) = 0 .......(1) \: \\ \\ or \: \: (3x + 5k)x + (k - 4)y + (1 - k) = 0 \\ \\ or \: \: (5k + 3)x + (k - 4)y = k - 1 \\ \\ or \: \: ( \frac{5k + 3}{k - 1} )x + ( \frac{k - 4}{k - 1} )y = 1 \\ \\ or \: \: \frac{x}{ \frac{k - 1}{5k + 3} } + \frac{y}{ \frac{k - 1}{k - 4} } = 1 \\ \\ intercepts \: are \: \frac{k - 1}{5k + 1} \: and \: \frac{k - 1}{k - 4}$

But intercepts are equal

$∴ \: \: \frac{k - 1}{5k + 3} = \frac{k - 1}{k - 4} \\ \\ ∴ \: \frac{1}{5k + 3} = \frac{1}{k - 4} \\ \\ ∴ \: 5k + 3 = k - 4 \\ \\ ∴ \: 4k = - 7 \\ \\ ∴ \: k = \frac{ - 7}{4}$

putting this value of k in (1) we get,

$(3x - 4y + 1) - \frac{7}{4} (5x + y - 1) = 0\\ \\ or \: 4(3x - 4y + 1) - 7(5x + y - 1) = 0 \\ \\ or \: 12x - 16y + 4 - 35x - 7y + 7 = 0 \\ \\ or \: \: - 23x - 23y + 11 = 0 \\ \\ or \: \: 23x + 23y - 11$