The equation of the straight line which passes through the point (3, 4) and whose intercept on y-axis is twice that on x-axis, is
2x + y = 10
Let the equation of the line be
xa+yb=1......(1)
Given: intercept on y-axis = 2 (intercept on x-axis)
i.e., b = 2a
So, equation (1) becomes
xa+y2a=1
⇒2x+y=2a……(2)
Since line (2) passes through the point (3, 4)
∴ 2×3+4=2a⇒a−5
Thus, the equation of the required line is
2x+y=10