the graph of equation y=ax+b,ab<>0,has equal x- and y-intercepts , the slope is equals to -1, how?

Question

<> means not equal

Answer 1

the graph of equation y=ax+b,ab ≠ 0,has equal x- and y-intercepts , the slope is equals to -1. How?

y = ax + b

When x = 0 (for y-intercept), y = b

When y = 0 (for x-intercept), ax + b = 0, x = -b/a

Since x-intercept = y-intercept

b = -b/a

So -1/a = 1 Thus a = -1

But a is the slope of the line (y = ax + b)

So the slope of the line is -1 ..... QED

(So any line y = b - x has the same intercepts on the x- and y-axes)

Answer 2

Assume that y = ax + b, for ab <> 0, has equal x and y intercepts. Algebraically, let's solve for the intercepts.

x-intercept: Make y = 0. Then 0 = ax + b, or -b = ax, or -b/a = x.

y-intercept: Make x = 0. Then y = a(0) + b, or y = b.

In order for the x-intercept and y-intercept to be equal, we have to equate them.

-b/a = b

Multiply both sides by a (which is a valid step, since a is non-zero), to get

-b = ab, and
ab + b = 0, which we can factor to get
b(a + 1) = 0, implying b = 0 or (a + 1) = 0.

b = 0, a = -1

This means we can plug in these values for the equation
y = ax + b. Plugging in a = -1 and b = 0,
y = -x + 0
y = -x, which clearly has a slope of -1.

Answer 3

y = -1*x + 1 has y-intercept of 1 and x-intercept of 1. Actually with a slope of -1, the x and y intercepts will always be numerically equal, as you have a series of isosceles right triangles.in both the 1st and 3rd quadrants.