Question

Writing to Learn \(x\) -and \(y\) -intercepts (a) Explain why \(c\) and \(d\) are the \(x\) -intercept and \(y\) -intercept, respectively, of the line $$\frac{x}{c}+\frac{y}{d}=1$$ (b) How are the \(x\) -intercept and \(y\) -intercept related to \(c\) and \(d\) in $$\frac{x}{c}+\frac{y}{d}=2 ?$$

Short answer

In equation \(\frac{x}{c}+\frac{y}{d}=1\), c and d are the x-intercept and y-intercept respectively. In equation \(\frac{x}{c}+\frac{y}{d}=2\), 2c and 2d are the x-intercept and y-intercept respectively.

Step-by-step solution

Analyze the first equation

The first equation is \(\frac{x}{c}+\frac{y}{d}=1\). To find the x-intercept, we need to set y equal to zero and solve for x. Doing this gives us \(\frac{x}{c}+\frac{0}{d}=1\), which simplifies to \(\frac{x}{c}=1\). Multiplying both sides by c, we find that x=c. Therefore, c is the x-intercept.

Solve for the y-intercept of the first equation

Similarly, to find the y-intercept, we need to set x equal to zero and solve for y. So our equation becomes \(\frac{0}{c}+\frac{y}{d}=1\), which simplifies to \(\frac{y}{d}=1\). Multiplying both sides by d, we find that y=d. Therefore, d is the y-intercept.

Analyze the second equation

The second equation is \(\frac{x}{c}+\frac{y}{d}=2\). Using the same method, if we set y to 0 and solve for x we get \(\frac{x}{c}=2\), which gives x=2c. Therefore, the x-intercept is twice the value of c.

Solve for the y-intercept of the second equation

Setting x equal to zero and solving for y in the second equation leads to \(\frac{y}{d}=2\), which simplifies to y=2d. Therefore, the y-intercept is twice the value of d.