Question

find the equation of the line whose sum and product of intercepts on
the coordinate axes are-1 and -12 r
respectively

Answer 1

Answer:

$y=\frac{3}{4}x+3$

Step-by-step explanation:

let the x-intercept be (x,0) and the y-intercept be (0,y).

If the sum of these two is -1, then we can form the equation;

$x+y=-1.......................(1)$

if the product of the intercepts is -12, then;

$x*y=-12.....................(2)$

we solve for the values of x and y using these two equations.

From eqn (1)

$x+y=-1\\x=-y-1\\x=-(y+1)$

substituting this in eqn (2) we get

$x*y=-12\\-(y+1)*y=-12\\y(y+1)=12\\y^2+y=12$

we solve this quadratic eqn using Complete the Square method

$y^2+y=12\\y^2+y+...=12+...\\y^2+y+0.5^2=12+0.5^2\\(y+0.5)^2=12+0.25\\(y+0.5)^2=12.25\\(y+0.5)=\sqrt{12.25}\\ y+0.5=3.5\\y=3.5-0.5\\y=3$

this implies that;

$x=-y-1\\x=-3-1\\x=-4$

Therefore our intercepts are (-4,0) and (0,3)

The gradient of the line connecting these two pints is

$m=\frac{y_2-y_1}{x_2-x_1}\\ \\m=\frac{3-0}{0--4}\\ m=\frac{3}{4}$

The equation of the line can be found as;

$y-y_1=m(x-x_1)\\y-0=\frac{3}{4}(x--4)\\ \\y=\frac{3}{4}x+3$