Question

In the xy-plane, the point (p,r) lies on the line with equation y=x+b, where b is a constant. The point with coordinates (2p,5r) lies on the line with equation y=2x+b. If p≠0, what is the value of r p ? A) 2/ 5 B) 34 C) 43 D) 2

Answer 1

Answer:

, r) lies on the line y = x + b, then

r = p + b…. (1)

If (2p, 5r) lies on the line y = 2x + b, then

5r = 2(2p) + b

or 5r = 4p + b…. (2)

Subtracting equation (1) from (2), we get

5r - r = 4p + b - p - b

4r = 3p or r/p = 3/4

Answer 2

Answer:

$The \: \: correct \: \: answer \: \: is  \: \: B,  \frac{3}{4}$

Step-by-step explanation:

Since the point ( p , r ) lies on the line with equation y = x + b, the point must satisfy the equation. Substituting p for x and r for y in the equation

$y=x+b \:  gives \:  r=p+b, or b = r−p.$

Similarly, since the point (2 p, 5 r) lies on the line with the equation y = 2x + b, the point must satisfy the equation. Substituting 2p for x and 5r for y in the equation.

$y=2x+b gives:$

$➠ \: 5r=2(2p) \: + \: b$

$➠ \: 5r \: = \: 4p \: + \: b$

$➠ \: \bold{b = 5r−4p.}$

Next, we can set the two equations equal to b equal to each other and simplify:

$b \: = \: r−p \: = \: 5r−4p$

$3p \: = \: 4r$

$Finally, \: \: to \: \: find \: \frac{r}{p} \: \: we \: \: need \: \: to \: \: divide \: \\ both \: \: sides \: \: of \: \: the \: \: equation \: \: by \: \:  p \: \:  and \: \: by \: \:  4:$

$3p \: = \: 4r$

$3 \: = \: \frac{4r}{p}$

$\frac{3}{4} = \: \frac{r}{p}$

$The \: \: correct \: \: answer \: \: is  \: \: B,  \frac{3}{4}$

If you picked choices A and D, you may have incorrectly formed your answer out of the coefficients in the point (2p,5r). If you picked Choice C, you may have confused r and p.

Note that while this is in the calculator section of the SAT, you absolutely do not need your calculator to solve it!