Question

Find the equation of the line, which passes through P (1, -7) and meets the axes at A and B respectively so that 4 AP - 3 BP = 0.

Answer 1

$\text{The equation of straight line is}y-y_1=m(x-x_1)\text{The line passes through (x, y) i. e., (1, - 7) and meets the axes at A and B}\Rightarrow \text{A point is (a, 0) and B is (0, b)}\frac{AP}{BP}=\frac{3}{4}\text{Using section formula}=\frac{l{x}_2+m{x}_1}{l+m},\frac{l{y}_2+m{y}_1}{l+m}l:m=3:4(a,0)\iff \left(x_1{y}_1\right),(0,b)\iff \left(x_2{y}_2\right)\Rightarrow 1=\frac{3\left(0\right)+4\left(a\right)}{3+4}\Rightarrow 1=\frac{4a}{7}\Rightarrow a=\frac{4}{7}-7=\frac{3\left(b\right)+4\left(0\right)}{3+4}\Rightarrow -7=\frac{3b}{7}\Rightarrow b=\frac{-49}{3}thenA(\frac{7}{4},0)B(0,\frac{-49}{3})\text{putting in (1)}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)y-0=\frac{\frac{-49}{3}-0}{0-\frac{7}{4}}(x-\frac{7}{4})y-0=\frac{49}{3}\times \frac{4}{7}(x-\frac{7}{4})y=\frac{28}{3}(x-\frac{7}{4})3y-28x+49=028x-3y=49$